To solve this problem, we need to calculate the energy released in the given nuclear reaction.$\\$The reaction is:$\\ ^7_3$Li$+ ^1_1$H$\to ^4_2$He$+ ^4_2$He$+ Q \\$First, we calculate the total binding energy of the reactants and products.$\\$Binding energy per nucleon of $^7_3$Li$$ is $5.60$ MeV.$\\$Number of nucleons in $^7_3$Li$$ is $7. \\$Total binding energy of $^7_3$Li$= 5.60 \times 7 = 39.2$ MeV.$\\$Binding energy per nucleon of $^1_1$H$$ is $0$ MeV (since it's a single proton).$\\$Total binding energy of $^1_1$H$= 0$ MeV.$\\$Total binding energy of reactants:$\\ 39.2 + 0 = 39.2$ MeV.$\\$Binding energy per nucleon of $^4_2$He$$ is $7.06$ MeV.$\\$Number of nucleons in each $^4_2$He$$ is $4. \\$Total binding energy of one $^4_2$He$= 7.06 \times 4 = 28.24$ MeV.$\\$Since there are two $^4_2$He$$ nuclei produced, the total binding energy of products is:$\\ 2 \times 28.24 = 56.48$ MeV.$\\$Total binding energy of products:$\\ 56.48$ MeV.$\\$The energy released $Q$ is the difference between the total binding energy of the products and the reactants:$\\ Q = 56.48 - 39.2 = 17.28$ MeV.$\\$Therefore, the value of energy $Q$ released is approximately $17.3$ MeV.$\\$This corresponds to Option 4.