Answer
17.35 MeV is released.
Work Step by Step
Calculate the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left(m_p+m(^{7}_{3}Li)-m(^{4}_{2}He)-m_{\alpha} \right)c^2$$
$$ =\left(1.007825u+7.016003u-4.002603u-4.002603u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=17.35MeV$$
The Q-value is positive. The reaction is exothermic, and releases 17.35 MeV of energy.