Below is an approach using elementary calculations.
We first find the number of subgroups of order $4$.
Denote $\operatorname{Gal}(L/\mathbb Q)=G$.
Note that the order of $G$ is $|H|\times|\operatorname{Gal}(L^H/\mathbb Q)|=20$, and every element in a subgroup of order $4$ has order dividing $4$. So we count the number of elements of order dividing $4$.
Let $\sigma\in G$ be an element of order dividing $4$. Since $\sigma$ is an automorphism, it must send $w$ to one of $w^i,\,i=1,\ldots,4$. If $\sigma(w)=w$, then its action is completely determined by $\sigma(\sqrt[5]3)=\sqrt[5]3w^k$ for some $k=0,\ldots,4$. So $\sigma$ has order dividing $5$, a contradiction unless $\sigma=1_G$.
Thus suppose $\sigma$ sends $w$ to $w^i,\,i=2,3,4$.
Suppose $\sigma(w)=w^i$ for some $i=2,3,4$, and $\sigma(\sqrt[5]3)=\sqrt[5]3\cdot w^k$, for some $k=0\ldots,4$. It follows that $\sigma^4(w)=w$, and
$$
\begin{align}
\sigma^2(\sqrt[5]3)&=\sqrt[5]3\cdot w^{k+ik}\\
\sigma^3(\sqrt[5]3)&=\sqrt[5]3\cdot w^{k+i(i+1)k}\\
\sigma^4(\sqrt[5]3)&=\sqrt[5]3\cdot w^{k+ik+i^2(i+1)k}=\sqrt[5]3\cdot w^{k(i+1)(i^2+1)}.
\end{align}
$$
It is easy to check that for $i=2,3,4$, we have $5\mid(i+1)(i^2+1)$, so $\sigma^4(\sqrt[5]3)=\sqrt[5]3$.
Consequently, an element $\sigma\in G$, which is not equal to $1_G$, is of order dividing $4$ if and only if it sends $w$ to $w^i$ for some $i=2,3,4$ and sends $\sqrt[5]3$ to $\sqrt[5]3\cdot w^k$ for some $k=0,\ldots,4$.
We shall now decide how many subgroups are formed by these elements. Denote the automorphism sending $w$ to $w^i$ and $\sqrt[5]3$ to $\sqrt[5]3\cdot w^k$ as $\sigma_{i,k}$.
Observe that the above calculation shows that $\sigma_{i,k}^2=\sigma_{i^2,(i+1)k}$. So $\sigma^2=1_G$ if and only if $i=4$. This also shows that if $i,j\ne4$ and $\sigma_{i,k}^2=\sigma_{j,\ell}^2$, then $i^2\equiv j^2\pmod5$ and $(i+1)k\equiv(j+1)\ell\pmod5$.
Then $i^2\equiv j^2\pmod5$ implies $i\equiv\pm j\pmod5$. And $\ell\pmod5$ is uniquely determined by the congruence equation $(i+1)k\equiv(j+1)\ell\pmod5$. So, for any $i=2,3$, and $k=0,\ldots,4$, there are exactly two $\sigma_{j,\ell}$ such that $\sigma_{i,k}^2=\sigma_{j,\ell}^2$. Since $\sigma_{i, k}$ is of order $4$, these two elements are exactly $\sigma_{i, k}$ and $\sigma_{i, k}^3$.
Thus we can divide the set of elements $\ne1_G$ of order dividing $4$ into $5$ mutually disjoint sets of the form $\left\{\sigma_{i, k},\sigma_{i,k}^2,\sigma_{i,k}^3\right\}$. We see immediately that they form (with $1_G$ adjoined) $5$ subgroups of $G$ of order $4$.
Finally, by an elementary calculation, we see that
$$
\sigma_{i, k}\cdot\sigma_{j,\ell}=\sigma_{ij,k+i\ell}.
$$
Clearly this is non-abelian.
Hope this helps.
