Answer
See answers.
Work Step by Step
a. Check the sign of the Q-value, using data from Appendix B.
$$Q=\left( m_n+m(^{24}_{12}Mg)-m(^{23}_{11}Na)-m_d \right)c^2$$
$$ =\left(1.008665u+23.985042u-22.989769u-2.014102u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=-9.468MeV$$
This is negative, so it requires energy to proceed. Yes, the 18.00 MeV of kinetic energy is enough to make the reaction happen.
b. The energy released is 18.00MeV - 9.468 MeV = 8.53MeV.