= Solution
Expand the state in the <energy eigenstate>[energy eigenstates] of $H_0$ as
$$
|\psi(t)\rangle=\sum_kc_k(t)e^{-iE_kt/\hbar}|k\rangle.
$$
Substitution into the <Time-dependent Schrödinger equation> and projection onto $\langle k|$ gives the exact <interaction picture> equations
$$
i\hbar\dot c_k(t)=\sum_jc_j(t)e^{i(E_k-E_j)t/\hbar}\langle k|\Delta(t)|j\rangle.
$$
In first-order <time-dependent perturbation theory>, put $c_j(t)=\delta_{j0}$ on the right-hand side and integrate from $0$ to $t$:
$$
\boxed{c_k(t)=\frac1{i\hbar}\int_0^t
\langle k|\Delta(t')|0\rangle e^{i(E_k-E_0)t'/\hbar}\,dt'.}
$$
For a uniform <electric field>, the <electric-dipole interaction> is $\Delta(t)=e\mathcal E_0ze^{-t/\tau}$ up to an irrelevant overall sign. The supplied <hydrogen atom> wavefunctions give
$$
\begin{aligned}
\langle210|z|100\rangle
&=\frac1{4\sqrt2\pi a_0^3}
\int_0^\infty\frac{r^4}{a_0}e^{-3r/(2a_0)}\,dr
\int\cos^2\theta\,d\Omega\\
&=\frac{2^8}{3^5\sqrt2}a_0.
\end{aligned}
$$
For $t\gg\tau$, the time integral is
$$
\int_0^\infty e^{-t/\tau+i\omega t}\,dt=\frac1{\tau^{-1}-i\omega}.
$$
The long-time transition fraction is consequently
$$
\boxed{|c_{210}(\infty)|^2
=\frac{2^{15}}{3^{10}}
\frac{e^2\mathcal E_0^2a_0^2}{\hbar^2(\omega^2+\tau^{-2})}.}
$$
Both $|100\rangle$ and $|200\rangle$ have even <parity>, whereas $z$ is odd, so the <electric-dipole selection rule> gives $\langle200|z|100\rangle=0$. Thus the requested fraction in $|200\rangle$ is \b[zero].
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