Construct an extension of $\mathbb{Q}(3^{\frac{1}{3}})$ by radicals.

Question

Let $\mathbb{Q}$ be the field of rational numbers. Suppose $$ \mathbb{Q}(3^{\frac{1}{3}}) \leq R \leq \bar{\mathbb{Q}}$$ where $R=\mathbb{Q}(3^{\frac{1}{3}},a,b)$, where $a^{n} \in \mathbb{Q}(3^{\frac{1}{3}})$ and $b^{m} \in \mathbb{Q}(3^{\frac{1}{3}},a)$ for positive integers $m,n$.

Is it possible to construct a field $K$ such that:

1. $R \leq K \leq \bar{\mathbb{Q}}$
2. $K$ is a finite Galois extension of $\mathbb{Q}(3^{\frac{1}{3}})$
3. $K$ is an extension of $\mathbb{Q}(3^{\frac{1}{3}})$ by radicals
4. $Gal(K/\mathbb{Q}(3^{\frac{1}{3}}))$ is solvable ?

How can I construct $K$? I tried to adjoin the $n, m$th roots of unity, but I cannot solve this. Thank You.