Calculate the binding energy of ${}_6{}^{12}C$. Also find the binding energy per nucleon. Given that mass of ${}_1{}^{1}H=1.0078 u,$ mass of ${}_0{}^{1}n=1.0087 u$ and mass of ${}_6{}^{12}C=12.00004 u$

One atom of  ${}_6{}^{12}C$ consists of $6$ protons, $6$ electrons and $6$ neutrons. The mass of the un-combined protons and electrons is the same as that of six ${}_1{}^{1}H$ atoms
(if we ignore the very small binding energy of each proton-electron pair).

Mass of six ${}_1{}^{1}H$ atoms $=6\times 1.0078=6.0468u$

Mass of six neutrons $=6\times 1.0087=6.0522u$

Total mass of the component particles $= 12.0990 u$

Mass defect is $∆m=0.0990 u$

Binding energy is $BE=\left(931\right)\left(0.099\right)=92 MeV$

Binding energy per nucleon is 

$\frac{BE}{A}=\frac{92}{12}=7.66 MeV$