(a) Fusion of $1 \mathrm{~kg}$ hydrogen in the sun $4_1^1{ }^1 \mathrm{H}+4 \mathrm{e}^{-} \longrightarrow{ }_2^4 \mathrm{He}+2 \mathrm{e}^{-}+2 v+6 \gamma+26.7 \mathrm{MeV}$ $\therefore 4$ atoms fuse to give $26.7 \mathrm{MeV}$, mass of each atom $=1$ u
$1 \mathrm{~kg}$ of atoms give $\frac{26.7}{4} \times 6.023 \times 10^{26} \mathrm{MeV}$ $=40.2 \times 10^{26} \mathrm{MeV}$.
(b) Fission of $1 \mathrm{~kg}$ of ${ }^{235} \mathrm{U}$ atoms. Each fision gives 200 $\mathrm{MeV}$ of energy.
$$
\begin{aligned}
&\therefore \quad 1 \text { atom } \longrightarrow 200 \mathrm{MeV} \text { of energy. } \\
&1 \text { mole }=235 \text { gm }=6.023 \times 10^{23} \text { atom } \\
&\therefore 6.023 \times 10^{23} \text { atom } \longrightarrow 200 \mathrm{MeV} \times 6.023 \times 10^{23} \mathrm{MeV} \\
&\therefore \quad 235 \mathrm{~g} \longrightarrow 200 \times 6.023 \times 10^{23} \mathrm{MeV} \\
&1 \mathrm{~kg}=10^3 \mathrm{~g} \longrightarrow \frac{200 \times 6.023 \times 10^{23}}{235} \mathrm{MeV} \times 10^3 \\
&\quad=5.12 \times 10^{26} \mathrm{MeV} \\
&\therefore \text { Ratio of } \frac{E_{\text {fussion }}}{E_{\text {fission }}}=\frac{40.2 \times 10^{26} \mathrm{MeV}}{5.12 \times 10^{26} \mathrm{MeV}} \simeq 8 .
\end{aligned}
$$