Algebraic extension of $\Bbb Q$ with exactly one extension of given degree $n$

If you choose a random $\sigma \in \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ and consider the fixed field $K\subset \bar{\mathbb{Q}}$ of $\sigma$, then $\bar{\mathbb{Q}}/K$ will be Galois. The Galois group $G$ will almost always be $\hat{\mathbb{Z}}$, the profinite completion of the integers, and this is a group that has exactly one finite index subgroup of each index. Then $K$ will solve your problem for each $n$. There will be infinitely many such fields.

The problem is that it is hard to write down any concrete element of the absolute Galois group (except complex conjugation, as lulu referred to), and then also hard to write down its fixed field. So I'm afraid this answer is very nonconstructive.

See here for details: https://mathoverflow.net/questions/273224/what-is-the-probability-of-generating-a-given-procyclic-subgroup-in-mathrmgal