What partial pressure of hydrogen gas is required in order for 0.00100 g of the gas to dissolve in 17.6 ml of pure water? the henry's law constant for hydrogen gas is 7.8 × 10–4 m atm–1?

There is a minor typo in this question. The value of 7.8x10^-4 m/atm is nonsense. However, looking up the Henry's constant for hydrogen gas, there is a value of 7.8x10^-4 m/(L*atm) which I will assume is the correct unit and the rest of the problem will be done with that in mind.  
First, determine what the molar concentration would be for 0.00100 g of H2 dissolved in 17.6 ml of water. Start with the atomic weight of hydrogen = 1.00794 g/mol. Molar mass of H2 is twice that, so 2 * 1.00794 = 2.01588 g/mol. So the number of moles of H2 we have is 0.001 g / 2.01588 g/mol = 0.000496061 mol. Finally, the molarity of the solution is 0.000496061 mol / 0.0176 L = 0.0281853 mol/L.  
Now we can use the equation 
Hcp = Ca/p 
where 
Hcp = Henry's constant 
Ca = Concentration in aqueous solution 
p = pressure  
So solve for p, substitute the known values, and calculate: 
Hcp = Ca/p 
p*Hcp = Ca 
p = Ca/Hcp  
p = (0.0281853 mol/L)/(7.8x10^-4 mol/(L*atm)) 
p = 36.135 atm  
Rounding to 3 significant figures gives p = 36.1 atmospheres.

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.