= Solution
Let $\alpha=\sqrt[5]3$ and let $\zeta$ be a primitive fifth root of unity. Then $L=\mathbb Q(\alpha,\zeta)$. The field degrees $[\mathbb Q(\alpha):\mathbb Q]=5$ and $[\mathbb Q(\zeta):\mathbb Q]=4$ both divide $[L:\mathbb Q]$, while adjoining the two generators gives $[L:\mathbb Q]\leq20$. Consequently $[L:\mathbb Q]=20$.
The <cyclotomic field> $\mathbb Q(\zeta)/\mathbb Q$ is Galois with cyclic group of order $4$. The <Fundamental theorem of Galois theory> gives a normal subgroup
$$
N=\operatorname{Gal}(L/\mathbb Q(\zeta)),\qquad |N|=5,\qquad G/N\cong C_4.
$$
Thus $N\cong C_5$ and \b[the chain] $\boxed{G\triangleright N\triangleright\{1\}}$ has abelian successive quotients, proving that $G$ is a <solvable group>. One may refine it by inserting the inverse image of the order-two subgroup of $C_4$, giving cyclic factors $C_2,C_2,C_5$.
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