= Solution
The only possible rational roots of the monic cubic are $1,-1$, and neither is a root. Thus $f$ is irreducible. The <discriminant of a depressed cubic> gives
$$
\Delta=-4\cdot2^3-27\cdot1^2=-59.
$$
It is not a rational square. By the permitted <cubic discriminant criterion>, the splitting field has <Galois group> $S_3$, degree six, and contains its quadratic subfield generated by $\sqrt\Delta$. Consequently
$$
\boxed{K=\mathbb Q(\sqrt{-59})\subseteq L,\qquad
[L:K]=3,\qquad\operatorname{Gal}(L/K)=A_3\cong C_3.}
$$
For a rational prime $r\ne59$, the reduction of $f$ is separable because its discriminant is a unit at $r$. Choose a finite extension of $\mathbb F_r$ in which that reduction splits. Part 1(a) lifts it to an <unramified extension> of $\mathbb Q_r$, and the <Hensel lemma> lifts all the simple roots into that extension. The local splitting field is therefore unramified over $\mathbb Q_r$. In particular all primes of $L/K$ above such $r$ are unramified.
At $59$, the supplied simple residue root $-28$ lifts to a root in $\mathbb Q_{59}$. The remaining factor is quadratic, so the local splitting field has degree at most two over $\mathbb Q_{59}$. It contains $\sqrt{-59}$, which is not a square in $\mathbb Q_{59}$ because its valuation is odd. It is therefore exactly $\mathbb Q_{59}(\sqrt{-59})$. This is also the completion of $K$ at its unique prime above $59$: the polynomial $T^2+59$ is <Eisenstein> there. Thus
$$
L_{\mathfrak P}=K_{\mathfrak p}=\mathbb Q_{59}(\sqrt{-59})
$$
for a prime $\mathfrak P$ above $\mathfrak p$. All the other primes above it give the same local conclusion by Galois conjugacy. The relative extension is therefore unramified at $59$ as well; indeed that prime splits completely in $L/K$.
The field $K$ has only complex places, so there is no infinite-place ramification. Hence $\boxed{L/K\text{ is unramified everywhere}}$.
We use the <Hilbert class field> theorem: for a number field $E$, its maximal everywhere unramified abelian extension $H_E$ is finite, contains every such extension, and satisfies $\operatorname{Gal}(H_E/E)\cong\operatorname{Cl}_E$, so $[H_E:E]=h_E$. The cyclic degree-three extension just constructed is contained in $H_K$. Therefore
$$
\boxed{3\mid h_K,\qquad h_K>1.}
$$
This is the <unramified cubic extension of Q of square root minus fifty-nine>; only divisibility of the class number, rather than its exact value, is needed here.
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