= Solution
Quantum statistics matter when thermal <energy> is comparable with a characteristic vibrational level spacing. The chain's natural <angular frequency> is $\omega_0=\sqrt{\kappa/m}=5.7471\times10^{12}\,\mathrm{s^{-1}}$, giving
$$
\boxed{T_Q\sim\frac{\hbar\omega_0}{k_B}=43.9\,\mathrm K.}
$$
Factors of order unity depend on which vibrational scale is used. The largest chain <phonon> <angular frequency> is $2\omega_0$, corresponding to $87.8\,\mathrm K$. An isolated two-atom bond has reduced <mass> $m/2$ and <angular frequency> $\sqrt2\omega_0$, corresponding to $62.1\,\mathrm K$. Thus quantum corrections become appreciable on the scale of several tens of kelvin, and all modes are classical only when $T$ is comfortably above the largest relevant vibrational scale. Below it, high-<frequency> <phonons> are progressively frozen out and <equipartition> fails.
For the isolated bond, the zero-temperature root-mean-square relative displacement is
$$
x_{\mathrm{zp}}=\sqrt{\frac{\hbar}{2(m/2)(\sqrt2\omega_0)}}\simeq0.0061\,\mathrm{nm}\simeq0.073\xi.
$$
It is small enough for a local bond expansion. The classical weakly anharmonic estimate in part d therefore has a useful parametric window
$$
\boxed{10^2\,\mathrm K\ll T\ll\text{a few }10^3\,\mathrm K,}
$$
where the lower condition means comfortably above the vibrational scale, rather than a sharp threshold. The model's upper bound concerns local anharmonicity and binding, not a prediction of the melting temperature.
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