= Solution
With the supplied sum, the second-order approximation is
$$
E_1\simeq\frac{h^2}{8ma^2}
-\frac{2048ma^4q^2\varepsilon^2}{h^2\pi^4}
(4.12\times10^{-3}).
$$
For an electron in the $1\,\mu\mathrm m$ box, $E_1^{(0)}=6.02\times10^{-26}\,\mathrm J$ and the correction is $-4.61\times10^{-27}\,\mathrm J$, giving \b[$5.56\times10^{-26}\,\mathrm J=3.47\times10^{-7}\,\mathrm{eV}$].
For $10\,\mu\mathrm m$, the formal result is approximately \b[$-4.61\times10^{-23}\,\mathrm J=-2.88\times10^{-4}\,\mathrm{eV}$]. It is not a credible approximation: even the minimum electric potential energy in that box is only $-5\,\mu\mathrm{eV}$, whereas this answer is much lower. The ground-to-first-excited-state mixing estimate increases from $0.160$ to $160$ when the length grows tenfold. The shift grows as $a^4$ while the level spacing decreases as $a^{-2}$, so the larger box is far outside the perturbative regime.
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