= Solution
Write the <Morse potential> relative to its minimum as $V(a+x)=D[e^{-2x/\xi}-2e^{-x/\xi}]$. Expanding one order beyond the harmonic term gives
$$
\boxed{V(a+x)=-D+\frac{D}{\xi^2}x^2-\frac{D}{\xi^3}x^3+O(x^4)
=-D+\frac\kappa2x^2-\frac{\kappa}{2\xi}x^3+O(x^4).}
$$
The negative cubic coefficient means that stretching is softer than compression. Thermal motion therefore has positive mean displacement even though the most probable minimum is at $x=0$.
For <thermal expansion of a weakly anharmonic bond>, classical <equipartition> gives $\langle x^2\rangle_0=k_BT/\kappa$. Vanishing mean restoring <force>, to first anharmonic order, requires
$$
0=\langle V'(a+x)\rangle=\kappa\langle x\rangle-\frac{3\kappa}{2\xi}\langle x^2\rangle+\cdots,
\qquad \langle x\rangle\simeq\frac{3k_BT}{2\kappa\xi}.
$$
For a bond free to expand under negligible applied tension, $L=\langle r\rangle\simeq a+\langle x\rangle$. Its <linear thermal expansion coefficient> is therefore
$$
\boxed{\alpha\sim\frac{k_B}{a\kappa\xi}\simeq3.9\times10^{-5}\,\mathrm{K^{-1}}.}
$$
Keeping the leading perturbative numerical factor gives $\alpha\simeq3k_B/(2a\kappa\xi)=5.9\times10^{-5}\,\mathrm{K^{-1}}$. The question only requires the scale, not that factor. The <thermal expansion> is positive; a purely harmonic approximation would give zero mean displacement and zero <thermal expansion>. The local <Taylor expansion> describes a bound bond and is not a globally stable cubic potential.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2018-morse-bond-asymmetry.png]
{title=<Morse potential> asymmetry and its local harmonic and cubic approximations}
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