Energy from fusion of $2\,\text{kg}$ hydrogen

The given fusion reaction is: $4\,{}^1_1H + 2e^- \rightarrow {}^4_2He + 2\nu + 6\gamma + 26.7\,\text{MeV}$

So, 4 hydrogen atoms release $26.7\,\text{MeV}$.

Number of hydrogen atoms in $2\,\text{kg}$

Since $1$ mole of hydrogen atoms has mass $1\,\text{g}$, $2000\,\text{g} = 2000\,\text{moles}$

Hence number of H atoms: $N_H = 2000N_A$

Number of fusion reactions possible: $\frac{2000N_A}{4} = 500N_A$

Therefore, $E_H = 500N_A \times 26.7\,\text{MeV}$

Energy from fission of $2\,\text{kg}$ of ${}^{235}U$

Molar mass of ${}^{235}U$ is $235\,\text{g/mol}$.

Number of moles in $2000\,\text{g}$: $\frac{2000}{235}$

Number of nuclei: $N_U = \frac{2000}{235}N_A$

Each nucleus gives $200\,\text{MeV}$.

Therefore, $E_U = \frac{2000}{235}N_A \times 200\,\text{MeV}$

Compute the ratio

$\frac{E_H}{E_U} = \frac{500N_A \times 26.7}{\left(\frac{2000}{235}N_A\right)\times 200}$

Cancel $N_A$: $\frac{E_H}{E_U} = \frac{500 \times 26.7 \times 235}{2000 \times 200}$

Since $\frac{500}{2000} = \frac{1}{4}$, $\frac{E_H}{E_U} = \frac{26.7 \times 235}{4 \times 200}$

$= \frac{26.7 \times 235}{800}$

$26.7 \times 235 = 6274.5$

Hence, $\frac{E_H}{E_U} = \frac{6274.5}{800} \approx 7.84$

Choose the nearest option

The closest option is: $\boxed{7.62}$

So, Option A is the best answer.

Why slight difference appears

The exact calculation using the given data gives about $7.84$. In exam problems, small differences can arise due to rounding/approximate atomic masses used. Among the options, $7.62$ is the nearest and hence correct.