 Solution

For $K=\mathbb{R}$, the answer is yes. A root of ${\Phi }_4(X)=X^2+1$ is $i$ or $-i$, so

For $K=\mathbb{Q}$, the answer is no. Every $\mathbb{Q}({\zeta }_n)/\mathbb{Q}$ is finite abelian Galois. If $\sqrt[3]{2}$ belonged to ${\mathbb{Q}}^{\mathrm{cyc}}$, then it would belong to some $\mathbb{Q}({\zeta }_n)$. The subextensions of an abelian Galois extension result would makeGalois. This is false: $X^3-2$ has two nonreal roots absent from the real field $\mathbb{Q}(\sqrt[3]{2})$. Hence ${\mathbb{Q}}^{\mathrm{cyc}}$ is a proper subfield of $\overline{\mathbb{Q}}$.

For $K={\mathbb{F}}_p$, the answer is yes. Given $m\ge 1$, takewhich is coprime to $p$. Every root of ${\Phi }_n$ in characteristic $p$ has exact multiplicative order $n$. The degreeis the least positive $d$ such that $n\mid p^d-1$. The value $d=m$ works, and no $d<m$ can work becauseThereforeEvery element algebraic over ${\mathbb{F}}_p$ lies in some finite field ${\mathbb{F}}_{p^m}$, soThese are respectively the maximal cyclotomic extension of the real numbers, the maximal cyclotomic extension of the rational numbers, and the maximal cyclotomic extension of a finite field.