= Solution
Suppose that $L/\mathbb Q$ were a <Finite Galois extension> with <Galois group> $S_3$ and unramified away from $7$. Its quadratic subfield is the fixed field of the alternating subgroup $A_3$. A quadratic field ramified only at $7$ must be $\mathbb Q(\sqrt{-7})$: the classification by <fundamental discriminants> shows that $-7$ is the only nontrivial quadratic discriminant supported at $7$. In particular, $7$ ramifies in this quadratic subfield.
Because $7$ does not divide $|S_3|=6$, all ramification at $7$ is <tame ramification>. Its <inertia group> is therefore cyclic. Its image in $S_3/A_3\simeq C_2$ is nontrivial because the quadratic subfield is ramified, so the inertia group is generated by a transposition and has order two; it cannot have order six because it is cyclic.
The <discriminant exponent of a tame Galois extension> is therefore
$$
v_7(d_L)=[L:\mathbb Q]\left(1-\frac1e\right)
=6\left(1-\frac12\right)=3.
$$
There is no other finite ramification, so $|d_L|=7^3=343$.
On the other hand, the <Minkowski bound for ideal classes> implies the <Minkowski lower bound for a number-field discriminant>
$$
|d_L|\geq
\left(\frac\pi4\right)^{2r_2}
\left(\frac{6^6}{6!}\right)^2.
$$
Since $0\leq r_2\leq3$, the right side is smallest at $r_2=3$ and is greater than $980$. This contradicts $|d_L|=343$, so no such extension exists.
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