1. Use binding energy per nucleon to find total binding energies

For deuteron $\left({}_1^2H\right)$:

• Number of nucleons $=2$
• Binding energy per nucleon $=1.1\,\text{MeV}$

So, total binding energy of one deuteron is $B_D = 2 \times 1.1 = 2.2\,\text{MeV}$

Since two deuterons react, $B_{\text{initial}} = 2 \times 2.2 = 4.4\,\text{MeV}$

2. Find total binding energy of helium nucleus

For helium nucleus $\left({}_2^4He\right)$:

• Number of nucleons $=4$
• Binding energy per nucleon $=7\,\text{MeV}$

So, total binding energy is $B_{He} = 4 \times 7 = 28\,\text{MeV}$

3. Energy released in the reaction

When lighter nuclei combine to form a more tightly bound nucleus, energy released is the increase in total binding energy: $Q = B_{\text{final}} - B_{\text{initial}}$

Thus, $Q = 28 - 4.4 = 23.6\,\text{MeV}$

4. Match with options

$23.6\,\text{MeV}$ corresponds to Option A.

Final Answer: $\boxed{23.6\,\text{MeV}}$