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A nuclear reactor operates using the fissionable material mentioned in the previous question. If it were to use $$540\text{ g}$$ of the material in a day, it's power would be (assume $$50\%$$ efficiency in conversion of energy released in the fission in to electrical energy)
Energy released per day $$= 540 \times 6.4 \times 10^{10} = 3.456 \times 10^{13}\text{ J}$$. Thermal power $$= \frac{3.456 \times 10^{13}}{86400} = 4 \times 10^8\text{ W}$$. With $$50\%$$ efficiency the electrical power is $$2 \times 10^8\text{ W} = 200\text{ MW}$$.
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