A town has a population of 1 million. The average electric power...

Total electric power required for the town is:

$P_e = 10^6 \times 300 \text{ W} = 3 \times 10^8 \text{ W}$

Given the efficiency of $25\%$, the total thermal power required is:

$P_{th} = \dfrac{P_e}{\eta} = \dfrac{3 \times 10^8}{0.25} = 1.2 \times 10^9 \text{ W}$

Total thermal energy required in $30$ days is:

$E = P_{th} \times t = 1.2 \times 10^9 \times (30 \times 24 \times 3600) = 3.1104 \times 10^{15} \text{ J}$

Energy released per fission of ^{235}\text{U}$is:<br><br>$E_f = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$<br><br>Number of fissions required per month is:<br><br>$N = \dfrac{E}{E_f} = \dfrac{3.1104 \times 10^{15}}{3.2 \times 10^{-11}} = 9.72 \times 10^{25}$<br><br>Mass of ^{235}\text{U}$ required is:

$m_{235} = \dfrac{N}{N_A} \times 235 \text{ g} = \dfrac{9.72 \times 10^{25}}{6.022 \times 10^{23}} \times 235 \text{ g} \approx 37924 \text{ g} = 37.92 \text{ kg}$

Since the uranium used is enriched to $3\%$ in ^{235}\text{U}$, the total mass of enriched uranium required is:<br><br>$M = \dfrac{m_{235}}{0.03} = \dfrac{37.92}{0.03} \approx 1264 \text{ kg}