Ramification in a splitting field

Edit: I didn't read the question thoroughly and was going to delete this answer since it does not use the hint supplied. But maybe I will leave it with the note that this does not use the given hint (I am happy to delete if OP wishes).

Consider an irreducible polynomial $g(x) \in \mathbb{Z}[x]$ such that a prime $p$ does not divide $disc(g(x))$. Let $\bar{g}[x] \in \mathbb{F}_p[x]$ be obtained from $g$ by reducing the coefficients. Since $p$ does not divide the discriminant of $g$ we have that $disc(\bar{g}(x)) \neq 0$, in particular $\bar{g}$ has distinct roots in $\mathbb{F}_p$.

Let $\mathfrak{p}$ lie above $p$ in $L$. Now consider the decomposition group $D_{\mathfrak{p}/p}$ and the inertia group $I_{\mathfrak{p}/p}$. We want to show that the inertia group is trivial (since this is the case if and only if $p$ is unramified in $L$.

The group $D_{\mathfrak{p}/p}$ acts on the roots of $g(x)$ faithfully (since these generate the extension of local fields $L_{\mathfrak{p}} / \mathbb{Q}_p$). But notice that the reduction map taking $$\{ \text{roots of } g(x) \} \to \{ \text{roots of } \bar{g}(x) \}$$ is injective (since both polynomials have distinct roots). Thus if $\sigma \in I_{\mathfrak{p}/p}$ (i.e., if $\sigma$ fixes the roots of $\bar{g}(x)$) then $\sigma$ must act trivially on the roots of $g(x)$ by the injectivity noted above. In particular $I_{\mathfrak{p}/p}$ is trivial.

The fact that $L/\mathbb{Q}$ is unramified away from primes dividing $D=\text{disc } \mathcal{O}_K$ is evident: $L$ is composition of different embeddings of $K$, each such embedding is unramified away from primes dividing $D$, so is their composition $L$.

Now we show that for $p\mid D$, $p$ has ramification index $2$ in $L$. Let $\alpha_i\in L$, $i=1,\cdots,5$ be roots of $f(X) = X^5-X+1$. By factoring $f$ modulo $p$, we see that there are exactly four distinct $\bar{\alpha}_i \in \bar{\mathbb{F}}_p$, say $\bar{\alpha}_1 = \bar{\alpha}_2$ and $\bar{\alpha}_1, \bar{\alpha}_3,\bar{\alpha}_4,\bar{\alpha}_5$ are distinct. Any inertia group above $p$ fixes $\alpha_3,\alpha_4,\alpha_5$, only non-trivial element for inertia group will be the swapping of $\alpha_1$ and $\alpha_2$. Therefore ramification index is $2$.

To compute the discriminant, you can use the discriminant formula for tame ramification. But a more elegant approach is to consider $F = \mathbb{Q}(\sqrt{D})$. Since every $p\mid D$ has ramification $2$ in $L$, $L/F$ is unramified at every finite prime. Note that $[L:F] = 60$, therefore $$|D_{L/\mathbb{Q}}| = |D_{F/\mathbb{Q}}|^{60} = 19^{60} 151^{60}$$