The BE/A for deuteron and an $\alpha -$ particle are $X_{1}and{X}_2$ respectively. The energy released in the reaction will be ${ }_1{H}^2+{}_1{H}^2\to {}_{2}H{e}^4$

Let's denote the BE/A for a deuteron as $X_1$ and the BE/A for an alpha particle as $X_2.$

In the given reaction: ${ }_1{H}^2+{}_1{H}^2\to {}_{2}H{e}^4$

The initial state has two deuterons (${ }_1{H}^2$) on the left-hand side, and the final state has two alpha particles (${ }_{2}H{e}^4$) on the right-hand side.

The total number of nucleons (protons + neutrons) on both sides of the reaction remains the same, as per the law of conservation of nucleons.

Since the energy released in a nuclear reaction is the difference between the total binding energies of the initial and final states, we can calculate the energy released (ΔE) as follows:

ΔE = (Initial binding energy) - (Final binding energy)

The initial binding energy is the binding energy of two deuterons, and the final binding energy is the binding energy of two alpha particles.

For the initial state: Binding energy of two deuterons = $4\times X_1=4X_1$

For the final state: Binding energy of two alpha particles = $4\times X_2=4X_2$

So, the energy released in the reaction is: 

$\Delta E = 4 \times X_1 - 4 \times X_2$

$\Delta E = 4\left(X_2 - X_1\right)$

Hence the correct answer is $4\left(X_2 - X_1\right)$.