\(=\frac{6.023 \times 10^{23}}{235} \times 2000\)

fission energy per atom \(=185 \mathrm{\,MeV}\)

\(\therefore\) Energy for \(2 \mathrm{\,kg}\) of fuel

\(=\frac{6.023 \times 10^{26} \times 2}{235} \times 185 \mathrm{\,MeV}\)

Power \(=\frac{\text { Energy released }}{\text { time }}\)

\((\) \(\because\) \( 1 \mathrm{\,MeV}=1.6 \times 10^{-13} \mathrm{\,J}, 30 \) days

\(=30 \times 24 \times 60 \times 60 \mathrm{\,sec}) \)

\(\therefore \) Power \(=\frac{6.023 \times 10^{26} \times 2 \times 185 \times 1.6 \times 10^{-13}}{235 \times 30 \times 24 \times 60 \times 60}\)

\(=\frac{3552 \times 10^{13}}{235 \times 3 \times 6 \times 6 \times 24 \times 10^{3}} \mathrm{\,W}\)

\(=\frac{3552 \times 10^{10}}{235 \times 3 \times 6 \times 6 \times 24}\)

\(=\frac{3552 \times 10^{4}}{235 \times 18 \times 6 \times 24} \mathrm{\,MW}=58.3 \mathrm{\,MW}\)