Answer
0.863 MeV.
Work Step by Step
The nuclear capture described is $^{7}_{4}Be+\;^{0}_{-1}e\rightarrow\;^{7}_{3}Li +\nu$.
Calculate the energy release from the difference in the masses, using data from Appendix B.
The mass of the captured $\beta^-$, which is an electron, is accounted for by adding 3 electrons to each side of the equation, then using the atomic masses.
$$E_{release}=\left( m(^{7}_{4}Be)-m(^{7}_{3}Li) \right)c^2$$
$$ =\left(7.016929u-7.016003u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ E_{release}=0.863\;MeV$$
