= Solution
If
$$
a=\frac12\kappa R^2\gg k_BT,
$$
then $e^{\beta a}$ is large and the wall correction is exponentially small:
$$
\boxed{\langle E\rangle\sim2k_BT}.
$$
This is exactly the <equipartition theorem> result from the two kinetic and two harmonic-potential quadratic terms.
If $a\ll k_BT$, use
$$
\frac a{e^{\beta a}-1}
=\frac1\beta-\frac a2+O(\beta a^2).
$$
Then
$$
\boxed{
\langle E\rangle
=k_BT+\frac{\kappa R^2}{4}
+O\left(\frac{\kappa^2R^4}{k_BT}\right)
\sim k_BT
}.
$$
The leading term is the kinetic energy. The particle is nearly uniform in the disk, so $\langle r^2\rangle\to R^2/2$ and $\langle V\rangle\to\kappa R^2/4$, much smaller than $k_BT$. The hard wall therefore suppresses the positional equipartition contribution, as in the <hard-wall correction to harmonic equipartition>.
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