= Solution
Use the intended <infinite square well> interpretation: the kinetic-energy operator on $(-a,a)$ with zero <Dirichlet boundary conditions>. The variational <energy> is most safely evaluated through its <quadratic form>,
$$
F[\phi]=\frac{\hbar^2}{2m}\frac{\int_{-a}^a|\phi'(x)|^2\,dx}{\int_{-a}^a|\phi(x)|^2\,dx}.
$$
The boundary term vanishes because the trial function vanishes at both walls. Its zero extension has <derivative> jumps there, so the form interpretation avoids assigning an unrestricted full-line second <derivative> to it.
Set $u=x/a$ and $b=\lambda a^2$. Apart from an irrelevant factor $a^2$, the trial function is $q_b(u)=(1-u^2)(1+bu^2)$. <Polynomial> <integration> gives
$$
\begin{aligned}
D(b)&=\int_{-1}^1q_b^2\,du=\frac{16}{315}(b^2+6b+21),\\
N(b)&=\int_{-1}^1(q_b')^2\,du=\frac8{105}(11b^2+14b+35).
\end{aligned}
$$
Hence the <polynomial Rayleigh-Ritz trial space for a one-dimensional box> has <energy>
$$
F(b)=\frac{\hbar^2}{ma^2}\frac{3(11b^2+14b+35)}{4(b^2+6b+21)}.
$$
The denominator is strictly positive, since $b^2+6b+21=(b+3)^2+12$. Differentiation gives
$$
\frac{ma^2}{\hbar^2}\frac{dF}{db}=\frac{3(13b^2+98b+21)}{(b^2+6b+21)^2}.
$$
The two stationary parameters are $b_\pm=(-49\pm4\sqrt{133})/13$. The smaller is a local maximum; the larger is a local minimum. Since the <energy> tends to $33\hbar^2/(4ma^2)$ as $|b|\to\infty$, the latter is the global minimum:
$$
\boxed{\lambda_* =\frac{-49+4\sqrt{133}}{13a^2}\simeq-\frac{0.220750}{a^2},\qquad E_{\rm upper}=\left(7-\frac{\sqrt{133}}2\right)\frac{\hbar^2}{ma^2}\simeq1.233718703\frac{\hbar^2}{ma^2}.}
$$
The upper stationary value is $(7+\sqrt{133}/2)\hbar^2/(ma^2)$. Exact roots give a slightly tighter bound than substituting the rounded hints.
\b[There are two qualifications to the printed operator and comparison <energy>.] For the box with <Dirichlet boundary conditions> and width $2a$, the actual <ground-state energy> is
$$
\boxed{E_0^{\rm box}=\frac{\pi^2\hbar^2}{8ma^2}\simeq1.233700550\frac{\hbar^2}{ma^2},}
$$
which lies just below the variational result. The coefficient $6/5$ printed as an exact <eigenvalue> is inconsistent with this box problem. Also, merely declaring the differential expression to be zero outside the interval does not impose hard-wall confinement or specify a self-adjoint domain. Read literally on a space admitting exterior-supported states, it has zero-energy states outside and does not have the claimed positive ground <energy>. The boxed variational result is the well-defined answer for the intended confined particle; a different full-line operator would need its domain and matching conditions supplied.
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