Let $P\in{\mathbb Q}[X]$ be an irreducible polynomial of degree $n\geq 3$, and let $\mathbb L$ be the decomposition field of $P$. Denote the Galois group of the extension ${\mathbb L}:{\mathbb Q}$ by $G$. Suppose that $G$ is solvable.
It is well-known then that any root $\alpha$ of $P$ is solvable by radicals, i.e. there is a tower of extensions ${\mathbb K}_0\subseteq {\mathbb K}_1 \subseteq \ldots \subseteq {\mathbb K}_r$ where ${\mathbb K}_0={\mathbb Q}$, $\alpha\in {\mathbb K}_r$ and for $1\leq i \leq r$, ${\mathbb K}_i$ is an elementary radical extension of ${\mathbb K}_{i-1}$ : ${\mathbb K}_i={\mathbb K}_{i-1}({\theta_i}^{m_i})$ for some $\theta_i\in{\mathbb K}_{i-1}$ and $m_i\geq 2$. Denote by $D$ the total degree of the extension ${\mathbb K}_r:{\mathbb Q}$.
Question. Is there a bound $f(n)$, depending only on $n$, such that in the tower construction above, we may take $D\leq f(n)$ ?
Heuristics for it : this is true for $n\leq 4$ : trivially $f(2)=2$, and we have $f(3)\leq 2\times 3\times 2$ (if we use Lagrange's method as in https://en.wikipedia.org/wiki/Cubic_equation#cite_note-36, we need $2$ for $j=e^{\frac{2\pi i}{3}}$, the third root of unity, and $3\times 2$ for the rest) and $f(4)\leq 2f(3)$ (since the method described in https://en.wikipedia.org/wiki/Quartic_equation#Galois_theory_and_factorization reduces solving a quartic to taking a square root and solving a cubic).
Heuristics against it : the proof I know of the fact that $\alpha$ is solvable by radicals when $G$ is solvable, relies on the Kronecker-Weber theorem. Now, if you look at the simplest case of this theorem, where you have to include a quadratic field ${\mathbb Q}(\sqrt{d})$ in a cyclotomic field $C_m={\mathbb Q}(e^{\frac{2\pi i}{m}})$, clearly a given $C_m$ can contain $\sqrt{d}$ for only finitely many square-free $d$'s, so that $m$ necessarily gets larger as $d$ gets larger.
A Galois extension is solvable iff it is at the top of a tower of abelian extensions (the definition of solvable Galois group)
Any abelian extension of degree $m$ is radical if the base field contains $\zeta_m$ (Kummer theory)
By induction on $m$, $\Bbb{Q}(\zeta_3,\zeta_4,\ldots,\zeta_m)/\Bbb{Q}$ is a radical extension, of degree $\le m!$
$L$ is the splitting field of a rational polynomial of degree $n$ so $[L:\Bbb{Q}]\le n!$
You assume that $Gal(L/\Bbb{Q})$ is solvable, so $L/\Bbb{Q}$ is a tower of abelian extensions, each of degree $\le n!$
thus $L(\zeta_3,\ldots,\zeta_{n!})/\Bbb{Q}(\zeta_3,\ldots,\zeta_{n!})$ is a radical extension of degree $\le n!$ and $$L(\zeta_3,\ldots,\zeta_{n!})/\Bbb{Q}$$ is a radical extension of degree $\le f(n)=(n!)! \cdot n!$