Answer
$$5.8 \times 10^{3} \mathrm{kg}
$$
Work Step by Step
Using the result of Eq. $43-6,$ we obtain
$$
M=\frac{m_{\mathrm{U}} E_{\text {total }}}{Q}$$$$=\frac{(235 \mathrm{u})\left(1.66 \times 10^{-27} \mathrm{kg} / \mathrm{u}\right)\left(15 \times 10^{9} \mathrm{W} \cdot \mathrm{y}\right)\left(3.15 \times 10^{7} \mathrm{s} / \mathrm{y}\right)}{(200 \mathrm{MeV})\left(1.6 \times 10^{-13} \mathrm{J} / \mathrm{MeV}\right)}$$$$=5.8 \times 10^{3} \mathrm{kg}
$$
