= Solution
Normalize the trial <quantum state> as $(|\phi_0\rangle+c|\phi_1\rangle)/\sqrt{1+c^2}$. With the phases used above, the off-diagonal position <matrix element> is $x_{01}=-32a/(9\pi^2)$. Define
$$
\Delta=\epsilon_1^{(0)}-\epsilon_0^{(0)}=3K,
\qquad W=eEx_{01}=-\frac{32eEa}{9\pi^2}.
$$
The diagonal perturbation <matrix elements> vanish by parity, so the <Rayleigh quotient> is
$$
\mathcal E(c)=\frac{\epsilon_0^{(0)}+2cW+c^2\epsilon_1^{(0)}}{1+c^2}.
$$
The stationarity equation is $W+\Delta c-Wc^2=0$. The minimum continuously connected to the unperturbed ground <energy eigenstate> has
$$
\boxed{c_{\min}=-\frac W\Delta+O(E^3)
=\frac{256meEa^3}{27\pi^4\hbar^2}+O(E^3).}
$$
Equivalently, diagonalize the <Hamiltonian operator> matrix in this two-dimensional trial space. Its lower <eigenvalue> is
$$
\mathcal E_{\min}=\frac{\epsilon_0^{(0)}+\epsilon_1^{(0)}}2
-\sqrt{\left(\frac\Delta2\right)^2+W^2}
=\epsilon_0^{(0)}-\frac{W^2}\Delta+O(E^4).
$$
Therefore
$$
\boxed{\mathcal E_{\min}=\frac{\pi^2\hbar^2}{8ma^2}
-\frac{8192me^2E^2a^4}{243\pi^6\hbar^2}+O(E^4).}
$$
This retains exactly the $k=1$ term of the full ground-state <second-order nondegenerate perturbation theory> sum. Every omitted term is negative, so the full quadratic correction is more negative than this restricted one. This is consistent with the <Rayleigh-Ritz variational principle>: minimizing over a two-state subspace gives an upper bound on the exact ground energy. The bound applies to the unexpanded two-state minimum; a truncated series is to be compared order by order.
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