1. Use binding energy difference

When two deuterons fuse: $2\, {}_1^2\mathrm{H} \rightarrow {}_2^4\mathrm{He}$

The energy released is the increase in total binding energy: $Q = B_{\text{final}} - B_{\text{initial}}$

2. Initial total binding energy

For one deuteron, binding energy per nucleon is given as $1.1\,\text{MeV}$. Since deuteron has $2$ nucleons, $B({}_1^2\mathrm{H}) = 2 \times 1.1 = 2.2\,\text{MeV}$

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

3. Final total binding energy

For ${}_2^4\mathrm{He}$, binding energy per nucleon is $7.0\,\text{MeV}$. Since helium-4 has $4$ nucleons, $B({}_2^4\mathrm{He}) = 4 \times 7.0 = 28.0\,\text{MeV}$

4. Energy released

$Q = 28.0 - 4.4 = 23.6\,\text{MeV}$

5. Check options

• A: $26.8\,\text{MeV}$ — incorrect
• B: $8.1\,\text{MeV}$ — incorrect
• C: $23.6\,\text{MeV}$ — correct
• D: $5.9\,\text{MeV}$ — incorrect

Therefore, the correct answer is: $\boxed{23.6\,\text{MeV}}$