Answer
5.515 MeV.
Work Step by Step
a. Column 6 of Appendix B shows that $^{24}_{11}Na$ is a $\beta^-$ emitter.
b. The decay described is $^{24}_{11}Na\rightarrow\;^{24}_{12}Mg +\;^{0}_{-1}e+\overline{\nu}$.
Calculate the energy release from the difference in the masses, using data from Appendix B.
The mass of the emitted $\beta^-$, which is an electron, is accounted for by adding 11 electrons to each side of the equation, then using the atomic mass of $^{24}_{12}Mg $ on the right hand side.
$$E_{release}=\left( m(^{24}_{11}Na)-m(^{24}_{12}Mg) \right)c^2$$
$$ =\left(23.990963u-23.985042u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ E_{release}=5.515MeV$$
If we assume that the magnesium and the neutrino have negligible KE, and assume zero mass for the neutrino, the energy release is the maximum kinetic energy of the emitted $\beta^-$.
$$ KE_{max}=5.515MeV$$