Two nuclei of mass number 3 combine with another nucleus of mass...

The nuclear reaction is given by:

$2 X^{3} + Y^{4} \rightarrow Z^{10}$

The total binding energy of the reactants is the sum of the binding energies of the two nuclei of mass number $3$ and the nucleus of mass number $4$.

Binding energy of reactants $= 2 \times (3 \times 5.6) + (4 \times 7.4)$

$= 33.6 + 29.6 = 63.2$ MeV

The total binding energy of the product nucleus of mass number $10$ is:

Binding energy of product $= 10 \times 6.1 = 61.0$ MeV

The change in mass energy $\Delta M c^2$ is equal to the difference in the total binding energies of the reactants and the product.

$\Delta M c^2 = |63.2 - 61.0| = 2.2$ MeV

Answer: $2.2$