Answer
5.702 MeV of energy is released.
Work Step by Step
Calculate the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left(m_{\alpha}+m(^{9}_{4}Be)-m(^{12}_{6}C)-m_{n} \right)c^2$$
$$ =\left(4.002603u+9.012183u-12.000000u-1.008665u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=5.702MeV$$
The Q-value is positive. The reaction is exothermic, and releases 5.702 MeV of energy.