Answer
0.019 MeV.
Work Step by Step
The decay described, considering the nuclei only, is $^{3}_{1}H\rightarrow\;^{3}_{2}He +\;^{0}_{-1}e+\overline{\nu}$.
Calculate the energy release from the difference in the masses, using data from Appendix B. The mass of the emitted $\beta$ particle is accounted for by adding one electron to each side of the nuclear reaction so we can use atomic masses. We see that the mass of the $^{3}_{2}He $ nucleus plus 2 electrons equals the atomic mass of $^{3}_{2}He$.
$$E_{release}=\left( m(^{3}_{1}H)-m(^{3}_{2}He) \right)c^2$$
$$ =\left(3.016049u-3.016029u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ =0.019MeV$$
