Answer
7.72 MeV
Work Step by Step
$\Delta m= [2(mass\,^{1}_{1}H)+(3-2)(mass\,^{1}_{0}n)]-(3.016030\,amu)$
$=[2(1.00783\,amu)+1(1.00866\,amu)]-(3.016030\,amu)$
$=0.00829\,amu$
This change in mass is converted to energy.
When 1 amu mass is converted, 931.5 MeV energy is obtained.
$\implies E_{b}=0.00829\times931.5\,MeV$
$=7.72\,MeV$
