Answer
(a) 92.2 MeV
(b) 7.68 MeV/nucleon
Work Step by Step
(a) $\Delta m=[m_{p}Z+(A-Z)m_{n}+m_{e}Z]-M$$=[(1.007276u\times6)+(6\times1.008665u)+(5.48578\times10^{-4}\,u\times6)]-12\,u=0.0989375\,u$
The complete annihilation of 1 u of mass releases 931.5 MeV of energy.
Therefore, the total binding energy of the C-12 nucleus is
$E_{b}=0.0989375\,u\times931.5\,MeV=92.2\,MeV$
(b) Average binding energy per nucleon=$\frac{E_{b}}{total\,number\,of\,nucleons}=\frac{E_{b}}{Mass\,number}=\frac{92.2\,MeV}{12}=7.68\,MeV/nucleon$
