= Solution
The rational-root test shows that $f=X^3-X-1$ is irreducible. Its discriminant is
$$
\Delta=-4(-1)^3-27(-1)^2=-23,
$$
which is not a square in $\mathbb Q$. The <Galois group of an irreducible cubic> therefore gives
$$
\operatorname{Gal}(L/\mathbb Q)\cong S_3.
$$
Quadratic subfields correspond to order-three <subgroups>. The unique such <subgroup> is $A_3$, and its fixed field is generated by the alternating product of root differences, whose square is $\Delta$. Thus the unique quadratic subfield is
$$
\boxed{\mathbb Q(\sqrt{-23})}.
$$
Let the roots be $\alpha,\beta,\gamma$ and choose
$\delta=(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)$, so $\delta^2=-23$. Put $\theta=\alpha+\delta$. A nonidentity even permutation moves $\alpha$ and fixes $\delta$, so it cannot fix $\theta$. An odd permutation negates $\delta$. If it fixes $\alpha$, it plainly does not fix $\theta$. If it sends $\alpha$ to $\beta$ and fixed $\theta$, then $\beta-\alpha=2\delta$. Cancelling the factor $\beta-\alpha$ in the expression for $\delta$ would give
$$
1=\mathord\pm2(\alpha-\gamma)(\beta-\gamma)
=\mathord\pm2f'(\gamma).
$$
But $f'(\gamma)=3\gamma^2-1$, so this would make $\gamma^2$ equal to $1/6$ or $1/2$, contradicting the irreducibility of its cubic minimal <polynomial>. The other transpositions are identical. Hence $\theta$ has trivial stabilizer, so
$$
[\mathbb Q(\theta):\mathbb Q]=6=[L:\mathbb Q],
\qquad
\boxed{L=\mathbb Q(\alpha+\sqrt{-23})}.
$$
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