The binding energy of deuteron ${(}_1{H}^2)$ is 1.15 MeV per nucleon and an alpha particle ${(}_2^{4}He)$ has a binding energy of 7.1 MeV per nucleon. Then, in the reaction ${ }_1^{2}H{+}_1^{2}H{\to }_2^{4}He+Q$ , the energy Q released is 

Binding energy of ${ }_1^{2}H=1.5\times$ number of nucleons = 1.5 × 2= 2.3 MeV. 
Total binding energy of reactants = 2.3+2.3=4.6 MeV. 
Binding energy of ${ }_2^{4}H=7.1\times$number of nucleons $=7.1\times 4=28.4$ MeV.
$\therefore Q=28.4-4.6=23.8$ MeV.