Q.How much mass of 235U is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV.

[!TLDR] Energy/day $=3000\text{MW}\times86400\text{s}=2.592\times10^{14}$ J; energy/fission $=202.79\text{MeV}=3.249\times10^{-11}$ J; fissions/day $=7.978\times10^{24}$; mass $= …

The total thermal energy needed per day is $E=P\times t=3000\times10^6\text{ W}\times86400\text{ s}=2.592\times10^{14}$ J. Each fission releases $202.79$ MeV $=202.79\times1.602\times10^{-13}\text{ J}\approx3.249\times10^{-11}$ J. The number of fissions needed per day is $N_{fissions}=\frac{E}{E_{fission}}=\frac{2.592\times10^{14}}{3.249\times10^{-11}}\approx7.978\times10^{24}$. Since each fission consumes one $^{235}U$ nucleus, the mass of $^{235}U$ requir …

Convert the required power to total daily energy, divide by the energy released per fission to get the number of fissions needed, then convert that atom c …

Forgetting to convert the average energy per fission from MeV to joules before dividing into the total energy (in joules), or forgetting to convert megawa …