= Solution
{c}
For one Cartesian displacement, the harmonic potential is $m_{\mathrm{Ar}}\omega_E^2x^2/2$. At $T\gg T_E$, the <equipartition theorem> assigns $k_BT/2$ to this quadratic potential term. Hence
$$
\boxed{\langle x^2\rangle=\frac{k_BT}{m_{\mathrm{Ar}}\omega_E^2}=\frac{\hbar^2T}{m_{\mathrm{Ar}}k_BT_E^2}.}
$$
This is a one-coordinate <mean squared displacement>, not the total three-dimensional displacement variance, which is three times larger in the isotropic model.
For comparison, the exact <Einstein mean-square displacement> is
$$
\langle x^2\rangle=\frac{\hbar}{2m_{\mathrm{Ar}}\omega_E}\coth\left(\frac{T_E}{2T}\right).
$$
Using $\coth y\simeq1/y$ recovers the requested high-temperature result. Its nonzero zero-temperature limit is the displacement contribution of the <zero-point energy>.
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