= Solution
For each coordinate, the horizontal graph position is
$$
r=\frac{T}{\theta_{\rm vib}}=\frac{(298)(0.695)}{\widetilde\omega}
=\frac{207.11}{\widetilde\omega},
$$
with $\widetilde\omega$ in inverse centimetres. Read $S/R$ from the supplied curve at these positions. Each $A_1$ frequency contributes one coordinate and each $E$ frequency contributes two. The graphical estimates are
$$
\begin{array}{c|c|c|c|c}
\widetilde\omega/\mathrm{cm}^{-1}&\text{degeneracy}&r&S/R\text{ per coordinate}&S/R\text{ including degeneracy}\\\hline
3034&1&0.0683&0&0\\
680&1&0.3046&0.17&0.17\\
363&1&0.5706&0.56&0.56\\
1220&2&0.1698&0.02&0.04\\
774&2&0.2676&0.12&0.24\\
261&2&0.7935&0.83&1.66
\end{array}
$$
Thus the listed frequency groups contribute approximately $0$, $1.4$, $4.7$, $0.33$, $2.0$ and $13.8\ \mathrm{J\,K^{-1}\,mol^{-1}}$, respectively, and
$$
\boxed{S_{\rm vib,m}\simeq2.67R\simeq22\ \mathrm{J\,K^{-1}\,mol^{-1}}.}
$$
The nine coordinates are consistent with $3N-6=9$ for a five-atom nonlinear <molecule>. The lowest-frequency doubly degenerate bend dominates the <vibrational entropy>; the C–H stretch is effectively frozen at this <temperature>.
As a numerical check on the graphical reading, the harmonic expression $S/R=x/(e^x-1)-\ln(1-e^{-x})$, with $x=1/r$, gives a total $22.10\ \mathrm{J\,K^{-1}\,mol^{-1}}$. This agrees with the precision of the supplied graph.
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