Answer
Mass defect = 0.04213 amu
Nuclear binding energy per nucleon = 5.606 MeV/nucleon
Work Step by Step
Mass defect=[Z(mass $^{1}_{1}H$)+(A-Z)(mass $^{1}_{0}n$)]-mass $^{7}_{3}Li$
=3(1.00783 amu)+4(1.00866 amu)- 7.016003 amu
=0.04213 amu.
Nuclear binding energy=$0.04213\,amu\times\frac{931.5\,MeV}{1\,amu}=39.24\,MeV$
Nuclear binding energy per nucleon=
$\frac{Binding\,energy}{Mass\,number}=\frac{39.24\,MeV}{7\,nucleons}=5.606\,MeV/nucleon$
