Answer
$ KE_{max}=0.782MeV$.
Work Step by Step
The decay described is $^{1}_{0}n\rightarrow\;^{1}_{1}p +\;^{0}_{-1}e+\overline{\nu}$.
Calculate the energy release from the difference in the masses, using data from Appendix B. The mass of the emitted electron is accounted for by using the atomic mass of $^{1}_{1}H$ on the right hand side.
$$E_{release}=\left( m(^{1}_{0}n)-m(^{1}_{1}H) \right)c^2$$
$$ =\left(1.008665u-1.007825u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ E_{release}=0.782MeV$$
If we assume that the proton and the neutrino have negligible KE, and assume zero mass for the neutrino, the energy release is the maximum kinetic energy of the electron.
$$ KE_{max}=0.782MeV$$
