Fuel used is half in 5 yrs
$$
\therefore \mathrm{N}=\frac{\mathrm{N}_0}{2}, \text { No. of atoms in } 1 \mathrm{gm} \text { of }{ }^{235} \mathrm{U}=\frac{6.023 \times 10^{23}}{235}
$$
Total energy of reactor $=1000 \mathrm{MW}=10^9 \mathrm{~J} / \mathrm{S}$.
Energy per fission $=200$
$$
\mathrm{MeV}=200 \times 1.6 \times 10^{-13} \mathrm{~J}=3.2 \times 10^{-11} \mathrm{~J}
$$
Now, no. of fission $/ \mathrm{sec}=\frac{\text { energy } / \mathrm{sec}}{\text { energy per fission }}$
$$
=\frac{10^9}{3.2 \times 10^{-11}}=\frac{1}{3.2} \times 10^{20}
$$
Reactor runs $80 \%$ time,
$\therefore$ Time for which reactor is working
$$
\begin{aligned}
&=\mathrm{t} \times 80 \%=5 \mathrm{yr} \times \frac{80}{100} \\
&=\frac{80}{100} \times 5 \times 3.15 \times 10^7 \mathrm{sec} \\
\therefore \quad & \text { No. of fission in this time } \\
&=\frac{80}{100} \times 5 \times 3.15 \times 10^7 \times \frac{1}{3.2} \times 10^{20} \\
&=4 \times 10^{27}=\frac{\mathrm{N}_0}{2} \text { (half fuel) } \\
\therefore \quad & \mathrm{N}_0=8 \times 10^{27} \text { atoms in fuel. }
\end{aligned}
$$
Now 235 gm of ${ }^{235} \mathrm{U}=6.023 \times 10^{23}$ atoms
$$
\begin{aligned}
\Rightarrow 8 \times 10^{27} \text { atoms }=& \frac{235}{6.023 \times 10^{23}} \times 8 \times 10^{27} \\
=\frac{1880}{6.023} \times 10^4 \mathrm{gm} &=312.1 \times 10^4 \mathrm{gm} \\
&=3121 \mathrm{~kg} .
\end{aligned}
$$