Assuming that about 200 MeV of energy is released per fission of${}_{92}\mathrm{U}^{235}$ nuclei, then the mass of ${\mathrm{U}}^{235}$ consumed per per day in a fission reactor of power 1 megawatt will be approximately

Power P of fission reactor, 

$\mathrm{P}={10}^6\mathrm{watt}= {10}^6 \mathrm{joule}/\sec$

Time =t = 1 day = $24 \mathrm{X} 36 \mathrm{x} {10}^2 \sec$

Energy produced , U = Pt

or   $$\begin{gathered} \mathrm{U} ={10}^6 \mathrm{x} 24 \mathrm{x} 36 \mathrm{x} {10}^2 \\ =24 \mathrm{x} 36 \mathrm{x} {10}^8 \mathrm{joule} \end{gathered}$$

Energy released per fission of ${\mathrm{U}}^{235}$

$=200 \mathrm{MeV} =32 \mathrm{x} {10}^{-12} \mathrm{J}$

Number of ${\mathrm{U}}^{235}$  atoms used

$$\begin{gathered} =\frac{24 \mathrm{x} 36 \mathrm{x} {10}^8}{32 \mathrm{x} {10}^{-12}}=27 \mathrm{x} {10}^{20} \\ \mathrm{Mass} \mathrm{of} 6 \mathrm{x} {10}^{23} \mathrm{atoms} \mathrm{of} {\mathrm{U}}^{235}=235\mathrm{g} \end{gathered}$$

$$\begin{gathered} \mathrm{Mass} \mathrm{of} 27 \mathrm{x} {10}^{20} \mathrm{atoms} \mathrm{of} {\mathrm{U}}^{235} \\ =\left(\frac{235}{6 \mathrm{x} {10}^{23}}\right)\left(27 \mathrm{x} {10}^{20}\right) \\ =1.058 \mathrm{g}=1 \mathrm{g} \end{gathered}$$