= Solution
The trial is even under reflection about the midpoint. The normalized box <eigenfunctions> satisfy $\phi_n(a-x)=(-1)^{n+1}\phi_n(x)$, so <orthogonality> between even and odd functions gives $c_n=0$ for even $n$. If the higher odd contributions $n\geq5$ are neglected, the <weighted mean> from the <quantum variational principle> becomes
$$
\widetilde E\simeq
\frac{|c_1|^2E_1+|c_3|^2E_3}{|c_1|^2+|c_3|^2}.
$$
Rearranging yields the <two-level variational contamination estimate>
$$
\boxed{\frac{|c_3|^2}{|c_1|^2}\simeq
\frac{\widetilde E-E_1}{E_3-\widetilde E}.}
$$
The required assumption concerns the neglected contributions to the <energy> as well as their small probabilities. Numerically the two-level estimate is $0.0016542$. It is not an exact identity: direct expansion of this parabolic trial gives $c_n\propto n^{-3}$ for odd $n$, hence the exact ratio is $1/729=0.0013717$. The omitted higher odd states account for the difference.
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