= Solution
<Chloroform> has the <C3v molecular point group>. Its three identical $^{35}\mathrm{Cl}$ atoms are interchanged by three proper rotations, so \b[$\sigma=3$]. The three mirror operations do not increase the <rotational symmetry number> to six. Isotopic identity matters for this count.
Using the supplied wavenumber form of $k_{\rm B}$,
$$
\theta_a=\theta_b=\frac{0.110}{0.695}=0.15827\ \mathrm K,\qquad
\theta_c=\frac{0.0593}{0.695}=0.085324\ \mathrm K.
$$
All are far below $298\ \mathrm K$, justifying the high-temperature nonlinear <rigid rotor> expression. Then
$$
q_{\rm rot}=\frac{\sqrt\pi}{3}
\left[\frac{298^3}{(0.15827)^2(0.085324)}\right]^{1/2}
\simeq6.5741\times10^4,
$$
and
$$
\boxed{S_{\rm rot,m}=R[\ln q_{\rm rot}+3/2]
\simeq104.71\ \mathrm{J\,K^{-1}\,mol^{-1}}.}
$$
The calculation treats the <molecule> as rigid and separates electronic, vibrational and nuclear-spin contributions from the requested rotational term.
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