When an atom of ${}_{92}{U}^{235}$ undergoes fission, about 200MeV of energy is released. If a reactor  using $U^{235}$ has an output of 700MW , the number of uranium  atoms cosumed in a day nearly

Power $\frac{Energy}{time}\Rightarrow Energy=Pt$

$hereP=700mw=7\times {10}^{8}W$and time t=24 hrs=86.400sec

$\therefore$Energy released $=7\times {10}^8\times 86,400J$

For one atom enery$=200MeV=200\times 1.6\times {10}^{-13}J$

$\therefore$ No of atoms consumed in one day $=\frac{7\times {10}^8\times 86,400}{200\times 1.6\times {10}^{-13}}\approx 1.9\times {10}^{24}$