By eqn. (10.2.3), the second virial coefficient of the gas with the given interparticle interaction would be

\[
\begin{aligned}
a_{2} & =-\frac{2 \pi}{\lambda^{3}}\left[\int_{0}^{D}-1 \cdot r^{2} d r+\int_{D}^{\infty}\left\{e^{\varepsilon(\sigma / r)^{6} / k T}-1 \right\} r^{2} d r \right] \\
& =\frac{2 \pi}{\lambda^{3}}\left[\frac{1}{3} D^{3}-\int_{r_{0}}^{\infty} \sum_{j=1}^{\infty} \frac{1}{j !}\left(\frac{\varepsilon \sigma^{6}}{k T r^{6}} \right)^{j} r^{2} d r \right] \\
& =\frac{2 \pi D^{3}}{3 \lambda^{3}}\left[1-\sum_{j=1}^{\infty} \frac{1}{(2 j-1) j !}\left(\frac{\varepsilon \sigma^{6}}{k T D^{6}} \right)^{j} \right]
\end{aligned}
\]

cf. eqn. (10.3.6). For the rest of the question, follow the solution to Problem 10.7.