A nuclear power plant makes 2400 MW of heat in its core, and its efficiency in making electricity…

Step 1. The electric power. $0.34 \times 2400\ \text{MW} = 816$ MW.

Step 2. The rate. $\dfrac{2.4 \times 10^{9}\ \text{W}}{3.204 \times 10^{-11}\ \text{J}} = 7.49 \times 10^{19}\ \text{s}^{-1}$.

Step 3. The uranium. In a year there are $7.49 \times 10^{19} \times 3.156 \times 10^{7} = 2.36 \times 10^{27}$ fissions. The moles are $\dfrac{2.36 \times 10^{27}}{6.022 \times 10^{23}} = 3925$ mol, and the mass is $3925\ \text{mol} \times 0.235\ \text{kg/mol} = 922$ kg.

Step 4. The coal. The heat in a year is $2.4 \times 10^{9}\ \text{W} \times 3.156 \times 10^{7}\ \text{s} = 7.57 \times 10^{16}$ J. The coal is $\dfrac{7.57 \times 10^{16}\ \text{J}}{3.0 \times 10^{7}\ \text{J/kg}} = 2.52 \times 10^{9}$ kg.