The binding energy per nucleon of  ${}_3{}^{7}Li$and ${}_2{}^{4}He$ Nuclei are 5.60 MeV  and 7.06 MeV respectively. In the nuclear reaction

${}_3{}^7\mathrm{Li}+{}_1{}^1\mathrm{H}\to {}_2{}^4\mathrm{He}+{}_2{}^4\mathrm{He}+Q$

the value of energy Q released is 

$\text{ Binding energy of }{}_3{}^7\mathrm{Li}\text{ nucleus }=7\times 5.60\mathrm{MeV}=39.2\mathrm{MeV}$

$\text{ Binding energy of }{}_2{}^4\text{ He nucleus }=4\times 7.06 \mathrm{MeV}=28.24 \mathrm{MeV}$

the reaction is

${}_3{}^7\mathrm{Li}+{}_1{}^1\mathrm{H}\to 2\left({}_2{}^4\mathrm{He}\right)+Q$

$\therefore Q=2\left(\mathrm{BE}\text{ of }{}_2{}^4\mathrm{He}\right)-\left(\mathrm{BE}\text{ of }{}_3{}^7\mathrm{Li}\right)$

$=2\times 28.24\mathrm{MeV}-39.2\mathrm{MeV}=56.48\mathrm{MeV}-39.2\mathrm{MeV}$

$=17.28\mathrm{MeV}\approx 17.3\mathrm{MeV}$