Answer
8.65 Megaton TNT is the rating of the Deuterium bomb
Work Step by Step
Out of 500 kg of Deuterium, only 30% of the mentioned amount was used. Hence the energy released can be given by
$$
E=N Q= \frac{0.300 M}{5 m_{2H}}
$$
Thus, the rating is given by the following calculation
$$
\begin{aligned}
R &=\frac{E}{2.6 \times 10^{28} \mathrm{MeV} / \mathrm{megaton} \mathrm{TNT}} \\
&=\frac{ 0.300 \mathrm{g} \times 500 \mathrm{kg} \times 24.9 \mathrm{MeV}_{}}{502.0 \mathrm{\mu g} \times 1.66 \times 10^{-27} \mathrm{kg} / \mathrm{uh} \times 2.6 \times 10^{28} \mathrm{MeV} / \mathrm{megaton} \mathrm{TNT}} \\
&=8.65 \mathrm{megaton} \mathrm{TNT}
\end{aligned}
$$
