The Binding energy per nucleon of Li 37 and He 24 nucleon are...

$L{i}^{7} 4 + {H 1}^{1} \to \left(H{e}^{4} 2\right) 2$
$BE$ of products$= ( \left(5.6MeV\right) \times 7 ) + 0$
$= 39.2MeV$
${E}_{i} = - 39.2MeV$
$BE$ of reactant$= \left(7.06\right) \times 4 \times 2$
$= 56.48MeV$
${E}_{f} = - 56.48MeV$

As nuclear energy decreases, some energy will be released.
${Q}_{release} = {E}_{i} - {E}_{f} = \left(- 39.2\right) - \left(- 56.48\right) = 17.28MeV$