Nucleus A is having mass number 220 and its binding energy per...

The total binding energy of nucleus $A$ is,

${E}_{A} = 220 \times 5 . 6 = 1232MeV$

The total binding energy of $B$ and $C$ combined is,

${E}_{B} + {E}_{C} = 105 \times 6 . 4 + 115 \times 6 . 4 = 1408MeV$

Therefore, the energy released per fission is,

$Q = {E}_{B} + {E}_{C} - {E}_{A} = 1408 - 1232 = 176MeV$