= Solution
Choose the $z$-axis along $\mathbf E$, so the perturbation is
$$
\Delta H=e|\mathbf E|z
$$
up to an irrelevant sign convention for the electron charge. By <second-order nondegenerate perturbation theory>,
$$
E_{100}^{(2)}
=e^2|\mathbf E|^2
\sum_{\alpha\ne100}
\frac{|\langle\alpha|z|1,0,0\rangle|^2}
{E_1-E_\alpha}.
$$
Parts (a) and (b) restrict the contributing bound states to
$$
|n,1,0\rangle,\qquad n\geq2.
$$
For hydrogen,
$$
E_n=-\frac R{n^2},
$$
and hence
$$
\frac1{E_1-E_n}
=\frac1{-R+R/n^2}
=\frac1R\frac{n^2}{1-n^2}.
$$
Therefore
$$
\boxed{
E_{100}^{(2)}
=\frac{e^2|\mathbf E|^2}{R}
\sum_{n=2}^{\infty}
\frac{n^2}{1-n^2}
|\langle n,1,0|z|1,0,0\rangle|^2
}.
$$
Every denominator is negative, so the discrete-state contribution lowers the ground-state energy.
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