Answer
See answers.
Work Step by Step
Start with the $\alpha$ emission. Calculate the energy release from the difference in the masses, using data from Appendix B.
$$E_{release}=\left( m(^{218}_{84}Po) -m(^{214}_{82}Pb) -m(^{4}_{2}He) \right)c^2$$
$$ =\left(218.008973u-213.999806u-4.002603u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ E_{release}=6.114MeV$$
Next analyze the $\beta^-$ emission. Calculate the energy release from the difference in the masses, using data from Appendix B. Assume that the neutrino mass is negligible. To balance the electrons, use the mass of the At atom.
$$E_{release}=\left( m(^{218}_{84}Po) -m(^{218}_{85}At) \right)c^2$$
$$ =\left(218.008973u-218.008695u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ E_{release}=0.259MeV$$