= Solution
We use the following standard facts about <cyclotomic fields>. For an odd prime $\ell$, the unique quadratic subfield of $\mathbb Q(\zeta_\ell)$ is
$$
\mathbb Q\left(\sqrt{(-1)^{(\ell-1)/2}\ell}\right).
$$
Also $\mathbb Q(\zeta_{\ell^r})$ has degree $\varphi(\ell^r)=\ell^{r-1}(\ell-1)$, and $\ell$ is totally ramified, with a unique prime and residue field $\mathbb F_\ell$. These are the <quadratic subfield of a prime cyclotomic field> and <total ramification in a prime-power cyclotomic field> results.
Take
$$
E_r=\mathbb Q(\zeta_{59^r})\quad(r\geq1).
$$
Since $59\equiv3\pmod4$, every $E_r$ contains $K=\mathbb Q(\sqrt{-59})$, through its subfield $E_1$. The unique prime above $59$ is totally ramified in $E_r/K$: the absolute ramification index is $58\cdot59^{r-1}$, the index in $K/\mathbb Q$ is $2$, and the relative index is therefore
$$
e(E_r/K)=29\cdot59^{r-1}=[E_r:K].
$$
The field $K$ has no real places. By part (b), $h_K\mid h_{E_r}$, and part (a) showed $3\mid h_K$. Thus
$$
\boxed{3\mid h_{E_r}\quad\text{for every }r\geq1.}
$$
The <ring of integers of a number field> is a <Dedekind domain>, and is a <principal ideal domain> exactly when its <ideal class group> is trivial. Hence none of these rings of integers is principal. Finally the degrees $58\cdot59^{r-1}$ strictly increase, so the fields are pairwise distinct. The explicit infinite family of <nonprincipal cyclotomic fields of conductor a prime power> is
$$
\boxed{\mathbb Q(\zeta_{59}),\ \mathbb Q(\zeta_{59^2}),\ \mathbb Q(\zeta_{59^3}),\ldots.}
$$
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