= Solution
The nonzero cubes modulo thirteen are $\{1,5,8,12\}$, which excludes three. Thus this cubic has no root in its base <field> and is irreducible. For a root $\alpha$, the extension $\mathbb F_{13}(\alpha)$ has degree three. The base <field> already contains the primitive cube <root of unity> $3$, since $3^3=1$ and $3\ne1$. Hence the other two roots are $3\alpha$ and $9\alpha$, already in this extension, making it the full <splitting field>.
The <Frobenius endomorphism> satisfies $\alpha^{13}=\alpha(\alpha^3)^4=3\alpha$, so it cycles all three roots. Therefore
$$
\boxed{[L:\mathbb F_{13}]=3,\qquad\operatorname{Gal}(L/\mathbb F_{13})\cong C_3.}
$$
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