[WuwuAndNillump]

The situation basically starts like this, and I've been going about it for 3 nights to no avail.

SITUATION:

Dextrose is a form of sugar (mixture of glucose, C₅H₁₂O₆, and water) and is usually injected into a patient through intravenous infusion. It is used to treat dehydration, low blood sugar, and insulin shock.

In disasters, especially during earthquakes, typhoons, and tsunamis, it is very common to have lacking supplies in medicines to treat mass injured patients with lack of means of transportation in delivering fresh supplies.

1.   You need to administer 5.00% mass-volume dextrose to a patient; however, the available concentrations are not expressed in mass-volumes, which of the following can be a substitute? Density of glucose = 1.56 g/mL. Density of water = 1g/mL.

Available substitute concentrations:
a.   0.298 M
b.   5.50x10^-3 mole fraction glucose

CHOICES:

I.  0.298 M, because they have the same ratio of the amount of glucose and amount of solution
Ii. 5.50 x10^-3 mole fraction glucose, because they have the same ratio of the amount glucose and amount of solution
Iii. both i and ii can be substitutes since they they have equivalent ratio of solute and solution with the 5 % mass-volume dextrose
Iv. not enough data is provided

EDIT: I've tried putting hypothetical volumes of dextrose using the given mass-percent, then used it to calculate Molarity, but I'm always a bit off.

[AWK]

The percentage concentration for both solutions can be easily calculated. For a molar concentration solution, you must assume a density equal to the density of water or find the exact value in the tables (~ 1.02).
Of course, both solutions do not have exactly 5% concentration, so some acceptable error must be assumed (for medical concentrations, an error below 10% is usually acceptable).
Solid glucose density is useless.

[WuwuAndNillump]

The percentage concentration for both solutions can be easily calculated. For a molar concentration solution, you must assume a density equal to the density of water or find the exact value in the tables (~ 1.02).
Of course, both solutions do not have exactly 5% concentration, so some acceptable error must be assumed (for medical concentrations, an error below 10% is usually acceptable).
Solid glucose density is useless.

I tried your suggestion and I got somewhere between 0.299 M and 0.300 M but never 0.298 M, is that still acceptable or is that point difference too far off to not be a plausible choice?

[AWK]

You should convert to a percentage:
I. molar concentration; or II. mole fraction

[MNIO]

Wow.  3 nights is a long time.  Let's see if I can give you a hand.

first off let's verify you have these definitions
  molarity (M) = moles solute / Liter solution
  molality (m) = moles solute / kg solvent
  mole fraction in liquid phase (χ) = moles of 1 component / total moles
  % mass / Volume in the medical field = g / 100mL

that last one is a bit ambiguous so let's make sure we're on the same page
  5.00% m/V = 5.00g / 100mL solution

***********
now you have an issue with the problem statement..
the problem statement says
  (1) dextrose (C5H12O6)
  (2) density of glucose = 1.56 g/mL
  (3) 0.298M
  (4) 5.50x10^-3 mole fraction glucose
  (I) 0.298M because they have the same ratio of glucose and volume solution
  (II) 5.50x10^-3 mole fraction glucose because it has the same ratio of glucose and volume solution
  (III) both I and II because they have the same ratio of solute and solvent as the 5.00% dextrose

notice a couple of things
  (A) dextrose and glucose are used interchangeably
  (B) dextrose has the formula C5H6O12.. that is INCORRECT

let's spend a second and review "dextrose" vs "glucose"
  glucose is the molecule C6H12O6.. C6.. not C5
  glucose exists as "d" and "l" enantiomers... (dextro and levo)
  the "d-glucose" aka "dextro-glucose" is commonly called "dextrose"

meaing
  the formula for dextrose is C6H12O6 with molar mass 180.16g/mol and the terms glucose and
  dextrose are used interchangeably for this question.  C5H12O6 has molar mass = 168.15 g/mol fyi

*********
let's do a quick conversion

  0.298 mol dextrose    180.16g dextrose     53.7g dextrose       1 L          5.37g
 ---------------------- x -------------------- = ----------------- x ---------- = -------- = 5.37% (m/V)
       1 L solution             1 mol dextrose         1L solution       1000mL      100mL

if we use the WRONG formula and wrong molar mass of dextrose, we get this calculation

  0.298 mol dextrose    168.15g dextrose     50.1g dextrose       1 L          5.01g
 ---------------------- x -------------------- = ----------------- x ---------- = -------- = 5.01% (m/V)
       1 L solution             1 mol dextrose         1L solution       1000mL      100mL

and we can see that with the wrong information, 0.298M glucose = 5.01% (m/V) dextrose

*******
Moving right along let's convert 5.50x10^-03 mole fraction glucose to % (m/V) glucose.  let's start by assuming a total # of moles of say... 1000

                   mole dextrose       moles dextrose
χ dextrose = ----------------- = ------------------ = 5.50x10^-3
                    (total moles)             1000mol

so that moles dextrose = 1000 mol * 5.50x10^-3 = 5.50
and mole water = 1000 mol - 5.50 mol = 994.5 mol H2O

next, let's calculate volume of each and then make the terrible assumption that the volumes are additive (which they aren't but it's an approximately starting point and seems to be where you're problem is leading you)

using the CORRECT molar mass for dextrose
                               5.50mol    180.16g     1mL
  volume dextrose = ---------- x ---------- x ------- = 635.2mL
                                  1            1 mol        1.56g

using the INCORRECT molar mass for dextrose
                               5.50mol    168.15g     1mL
  volume dextrose = ---------- x ---------- x ------- = 592.8mL
                                  1            1 mol        1.56g

The volume of water is
                           994.5mol    18.02g        1mL
  volume water = ----------- x ---------- x ------- = 17921mL
                                 1            1 mol        1g

total volume = 18556mL (correct mw for dextrose) or 18514mL (incorrect mw for dextrose)

giving this % m/V of dextrose using CORRECT mw of dextrose
    5.50mol     180.16g    100   (5.50 / 18556 * 180.16 * 100) g dextrose
  ------------ x --------- x ---- = ---------------------------------------------- = 5.34% (m/V)
    18556mL    1 mol       100                 100mL solution

and this % m/V of dextrose using INCORRECT mw of dextrose
    5.50mol     168.15g    100   (5.50 / 18514 * 168.15 * 100) g dextrose
  ------------ x --------- x ---- = ---------------------------------------------- = 5.00% (m/V)
    18514mL    1 mol       100                 100mL solution

***********
meaning...
.. IF we use the wrong formula for dextrose.... choice III is the correct choice
.. but if we use the correct formula for dextrose... none of the choices are correct.