Answer
$$226 \mathrm{\ MeV} $$
Work Step by Step
$\text { Using Table } 37-3,$ the energy released in this fission process is
$Q=\left(m_{\mathrm{U}}+m_{n}-m_{\mathrm{Ce}}-m_{\mathrm{Ru}}-10 m_{e}\right) c^{2} $
$\quad=(238.05079 \mathrm{u}+1.00867 \mathrm{u}-139.90543 \mathrm{u}-98.90594 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u})-10(0.511 \mathrm{MeV}) $
$\quad=226 \mathrm{MeV} $
