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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 givesIt 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

For a rational prime $r\ne 59$, 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. Thusfor 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 $\mathrm{Gal}(H_E/E)\cong {\mathrm{Cl}}_E$, so $[H_E:E]=h_E$. The cyclic degree-three extension just constructed is contained in $H_K$. ThereforeThis 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.