Answer
$$ 7.31 \mathrm{\ MeV} \text { per nucleon. } $$
Work Step by Step
if we look at appendix $\mathrm{F}$ and/or $\mathrm{G},$
we can find that: $Z=107$ for bohrium, so this isotope has
$$N=A-Z=262-107=155 \mathrm{\ neutrons}$$ Thus,
$\begin{aligned} \Delta E_{\mathrm{bc} \mathrm{n}} &=\frac{\left(Z m_{\mathrm{H}}+N m_{n}-m_{\mathrm{Bh}}\right) c^{2}}{A} \\ &=\frac{((107)(1.007825 \mathrm{u})+(155)(1.008665 \mathrm{u})-262.1231 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u})}{262} \\ & \mathrm{which} \text { yields } 7.31 \mathrm{MeV} \text { per nucleon. } \end{aligned}$
