Here's how I approached it, no idea if it is correct (but you seem desperate so I thought I'd throw this out there since I worked out something for it at least

.

I first converted the given 187 mL of O

_{2}/g*hr into moles of O

_{2}/g*hr as follows:

PV=nRT

n = (1 atm)(187 cm

^{3})/(82.058 cm

^{3} atm mol

^{-1} K

^{-1})(273K) = .008347531 mol O2/g*hr

Next, I assumed that all the energy being oxidized was produced by burning glucose only. Then:

6O

_{2} + C

_{6}H

_{12}O

_{6} 6CO

_{2} + 6H

_{2}0 + 36 ATP

So now we can get the ATP pool produced/g*hr:

(.008347531 mol O2/g*hr)(36 mol ATP/6 mol O2) = .050085185 mol ATP/g*hr or 50 mmol ATP/g*hr

If I am interpreting this right this is the amount of ATP that needs to be produced from oxygen each hour per gram of flight muscle (it gets used up to replenish the 7 mmol requirement multiple times in 1 hour). If there is no more oxidative phosphorylation the amount produced becomes zero which means the fly craps out when its residual supply of 7 mmol runs out on its own which would take:

(

~~g~~*hr/50

~~mmol ATP~~)(7

~~mmol ATP~~/

~~g~~) = .14 hr or 8.4 minutes