A nucleus with mass number 242 and binding energy per nucleon as 7...

Binding energy is given by $E = ∆ m{c}^{2}$

where $∆ m$ is the mass defect.

The energy per nucleon of the nucleus having mass number $242$ is $7 . 6MeV$.

The initial binding energy is,

$BE = 242 \times 7 . 6MeV$

The energy per nucleon of the nucleus with mass number $121$ is $8 . 1MeV$.

Therefore, binding energy is

$BE ' = 2\left(121 \times 8 . 1MeV\right)$

The gain in the binding energy is

$BE ' - BE = ( 8 . 1 – 7 . 6 ) \times 242MeV = 121MeV$

$= 121MeV$