I figured out (a) but I am having trouble with (b):
(a) What is the efficiency of an out of condition professor who 2.10e5 J of useful work while metabolizing 500 kcal of food energy?
(b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 25%?


2. Relevant equations
OE + Wnc = OEf
η = Work_out / Work_in
Work_in = η * work_out



3. The attempt at a solution
For (a) I converted the calories to Joules
1 kcal = 4184 J
500 kcal (4184J/1 kcal) = 2.1e6 J

η = W_out/W_in
η = (2.10e5J) / (2.10e6J)
η = 10%

(b) I assume I could input the 25% into the equation and work backwards but it's not giving me a correct anwser. I've tried the following:

0.25 = (2.10e5J) / (kcal)
kcal = (0.25)(2.10e5J)
kcal = 52500J
Convert the Joules to kcal and get something like 12kcal

But that's wrong also. Any guidance would be greatly appreciated.

I wrote my equation wrong and got it now.

Should be:

0.25 = (2.10e5) / (kcal)

kcal = (210e5) / (I had times) (0.25)

kcal = 840e3 J

J to kcal    1kcal = 4184 J

 840e3 / 4184 = 201 kcal

0.25 = (2.10e5J) / (kcal)
kcal = (0.25)(2.10e5J)

This is incorrect.  If "0.25= (2.10 e5J)/(kcal)" then multiply both sides by (kcal) to get 0.25(kcal)= (2.10 e5J) and divide both sides by 0.25: kcal= (2.10e5 J)/(0.25)= 8.50e^5= 850000 kcal.