Explanation

(c) $\ _{1}H^{2}\ + \ \ _{1}H^{2}\ \rightarrow \ _{2}He^{3}\ + \ \ _{0}n^{1}\ + Q(energy)$
$\because$ 2 molecules of deuterium are fused, then released energy = Q
Hence, energy released per molecule = $\frac{Q}{2}$
Now, we know that number of molecules in one mole
$\ = 6.02\ \times \ 10^{23}$
Hence number of molecules in two moles = $2\ \times \ 6.02\ \times \ 10^{23}$
Hence, energy released when two mole of deuterium are fused $\ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{Q}{2} \times 2 \times 6.02 \times 10^{23} = Q \times 6.02 \times 10^{23}$