Just wanted to check my work here.
$x^6 - 4 = (x^3 - 2)(x^3+2).$ The immediate roots are $a=2^{1/3}$ and $b=-2^{1/3}.$ Other roots will be $\omega a, \omega^2a, \omega b, \omega^2b$ for $\omega = e^{2\pi i / 3}.$ The field $E= \mathbb{Q}(2^{1/3}, \omega)$ is the smallest field that contains all of these roots, it has dimension $9,$ because $|E : \mathbb{Q}| = |\mathbb{Q}(\omega): \mathbb{Q}(2^{1/3})||\mathbb{Q}(2^{1/3}): \mathbb{Q}| = 3 \cdot 3.$
You are close. You correctly found that the splitting field of $x^6-4$ is $E=\mathbb{Q}(\omega,a)$. So you want to calculate $|E:\mathbb{Q}(a)||\mathbb{Q}(a):\mathbb{Q}|$. However, $|\mathbb{Q}(\omega,a):\mathbb{Q}(a)| \neq 3$. One way to calculate this is to find the minimal polynomial of $\omega$ over $\mathbb{Q}(a)$. It turns out that it's $x^2 +x +1$ (The same as it is over $\mathbb{Q}$). Hence, $|\mathbb{Q}(\omega,a):\mathbb{Q}(a)|=2$.
Another way to see that the splitting field has degree 6 is to show that $\mathbb{Q}(\omega,a)=\mathbb{Q}(\sqrt{-3},a)$.