Answer
2.822 MeV.
Work Step by Step
The decay described is $^{60}_{27}Co\rightarrow\;^{60}_{28}Ni +\;^{0}_{-1}e+\overline{\nu}$.
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 27 electrons to each side of the equation, then using the atomic mass of $^{60}_{28}Ni $ on the right hand side.
$$E_{release}=\left( m(^{60}_{27}Co)-m(^{60}_{28}Ni) \right)c^2$$
$$ =\left(59.933816u-59.930786u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ E_{release}=2.822MeV$$
If we assume that the nickel and the neutrino have negligible KE, and assume zero mass for the neutrino, the energy released is the maximum kinetic energy of the emitted $\beta$.