Ok, I interpret that to mean they want the number of moles of hydrogen atoms.

So, if you know there are 0.236 moles of (NH2)2CO and you know there are 6.02 x 1023 molecules of (NH2)2CO in each mole of (NH2)2CO, and you know how many atoms of hydrogen are in each molecule of (NH2)2CO, all you need left is how many atoms of hydrogen are in each mole of hydrogen.

Here's the train of thought:

[tex]\frac {\text{0.236 moles (NH2)2CO}}{} \text{ X } \frac{6.02\text{x}10^{23}\text{ molecules (NH2)2CO}}{\text{1 mole (NH2)2CO}} \text{ X } \frac{\text {4 atoms of hydrogen}}{\text{1 molecule (NH2)2CO}} \text{ X } \frac{\text{1 mole of hydrogen atoms}}{6.02\text{x}10^{23}\text{ atoms of hydrogen}}[/tex]

Usually we take a conceptual shortcut and just realize that there are 4 moles of hydrogen atoms 1 every one mole of (NH2)2CO molecules, but I wanted to write it all out so you can see where the logic lies.

This is the kind of unit conversion you need to become proficient at in order to be successful in general chemistry.

Now you should have everything - just multiply the #s to get the correct answer. Note that from here you have access to all kind of information: how many grams of hydrogen are in 14.2 grams of (NH2)2CO, what the percentage by mass of hydrogen is in the molecule, and so forth. But it all starts with unit conversions and understanding stoichiometry.