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Pick some $\gamma\in L\setminus\mathbb Q$ which is not a square and let $\gamma_1,\ldots,\gamma_n$ be its orbit under $\operatorname{Gal}(L/\mathbb Q)$. Then it is an easy exercise in Galois theory to show that:

1. $L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})/L$ is Galois with abelian Galois group $\subseteq(\mathbb Z/2\mathbb Z)^n$.
2. $L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})/\mathbb Q$ is Galois with nonabelian Galois group $\subseteq\operatorname{Gal}(L/\mathbb Q)\ltimes(\mathbb Z/2\mathbb Z)^n$.