Determining battery run time from Ah and MJ?
Rather than give you a direct answer I will give a few pointers. You can add your calculations into your question to show your progress.


  Two 100 ohm resisters are in parallel across a battery. The battery can store 240 MJ of energy and is rated 240 Ah. The battery depletes fully after two hours. 


That's all fine but notice that we haven't been given the voltage. 


  What is the current flowing through one resistor.


Since they haven't given us the voltage it appears that this is designed to make you think.


  \$ P=RI^2 \$ 


This is normally written slightly differently as \$ P = I^2R \$ which might be a little more recognizable. Of these three variables we only know \$ R \$ at this time so we get \$ P=100 \text Ω *I^2 \$.


  \$ (240*10^6\ \text J)/2=I^2 * 100 * 7200 \$


There's a line of explanation missing here and they have shown one of the units, \$ J \$, but not the others. (Can you add in the other three units?)

Now since this equation has the number of joules on the left side we know that it is an energy equation and we know that \$ energy = power \times time \$.

Does that help?



Update after comment:

I didn't complete the calculations when I wrote the original answer. I now realise that you have been supplied conflicting information.


  I did 240Ah/2=120A (i.e. battery supplying 120A for two hours). Each resistor gets half this amount, hence the current through each resistor is 60A.


Correct. So the energy consumed is 120 A into 50 &Omega; for two hours giving us \$ E = I^2R \times t = (120 \text A)^2 \times 50 \text Ω \times 7200 \text s = 5,184 \ \text {MJ} \$ and not 240 MJ.

Working from the 240 MJ specified, and adding the units into the equation we get \$ (240*10^6\ \text {(J)})/2=I^2 \text {(A)}* 100 {(\Omega)} * 7200 \text {(s)} \$ so $$ I = \sqrt{\frac {240 \times 10^6 \text J}{2 \times 100 \text Ω \times 7200 \text s} } = 12.9 \ \text A $$
So, their calculated answer is at least correct if you use the energy rating but not the Ah rating.

Your question wrecked my head for a while until I realised that we are trying to resolve \$ E = I^2Rt \$ when we've been given E (240 MJ), R (50 &Omega;), t (7200 s) and you quickly calculated I(120 A). The equation is over-constrained or, to put it another way, there are conflicting constraints.



As an exercise you could try removing the resistor constraint and see what resistance value would satisfy the energy, Ah and time constraints