The energy released in the fusion of $2\ \text{kg}$ of hydrogen deep in the sun is $E_H$ and the energy released in the fission of $2\ \text{kg}$ of $^{235}\text{U}$ is $E_U$. The ratio $\dfrac{E_H}{E_U}$ is approximately: (Consider the fusion reaction as $4\,{}^1_1\text{H} + 2e^- \to {}^4_2\text{He} + 2\nu + 6\gamma + 26.7\ \text{MeV}$, energy released in the fission reaction of $^{235}\text{U}$ is $200\ \text{MeV}$ per fission nucleus and $N_A = 6.023\times10^{23}$)
Answer: (A) 7.62
Hydrogen: $2\ \text{kg} = 2000\ \text{mol}$, i.e. $2000N_A$ atoms, and every 4 atoms release $26.7\ \text{MeV}$:
$$E_H = \frac{2000N_A}{4}\times26.7 = 13350\,N_A\ \text{MeV}$$
Uranium: $\dfrac{2000}{235}N_A$ nuclei, each releasing $200\ \text{MeV}$:
$$E_U = \frac{2000}{235}\times200\,N_A \approx 1702\,N_A\ \text{MeV}$$
$$\frac{E_H}{E_U} \approx \frac{13350}{1702} \approx 7.8$$
The closest option is 7.62.
Solution by Sreeraj P, M.Sc Physics