Your result for the splitting field itself looks good.$^{*}$
Here's one way to complete the degree computation. Let $L:=\mathbb{Q}(\sqrt[6]{20})$. As you say, $[L:\mathbb Q]=6$ and $$\sqrt{5}=\frac{1}{2}(\sqrt[6]{20})^3 \in L.$$ If $\sqrt{2}\in L$, then $\mathbb Q(\sqrt{2},\sqrt{5})\subset L$. I claim that
$$[\mathbb Q(\sqrt{2},\sqrt{5}):\mathbb Q]=4. \,\,\,\, (*)$$
Given this, $\mathbb Q(\sqrt{2},\sqrt{5})\subset L$ is impossible because $4$ doesn't divide $[L:\mathbb Q]$. Hence, $(*)$ implies that $d_2=2$.
To prove $(*)$, we could appeal to this general result, which implies
$$[\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_n}):\mathbb Q]=2^n$$ for $p_i$'s distinct primes. However, for $n=2$, the argument is simple. If $\sqrt{2}=a+b\sqrt{5}$ for $a,b \in \mathbb Q$, then $(a+b\sqrt{5})^2 \in \mathbb Q$. This implies $\sqrt{5} \in \mathbb Q$ unless $b=0$, in which case we have $\sqrt{2} \in \mathbb Q$. Both are impossible, so $\sqrt{2} \notin \mathbb Q(\sqrt{5})$ and $[\mathbb Q(\sqrt{2},\sqrt{5}):\mathbb Q(\sqrt{5})]=2$.
$^{*}$ For posterity, a note on computing the splitting field: the roots of the polynomial are $\pm \sqrt{2}$ and $\zeta_6^n \sqrt[6]{20}$, where $\zeta_6$ is a primitive $6$-th root of unity. To determine the splitting field $K$, first note that polynomial splits over $\mathbb Q( \sqrt{2},\sqrt[6]{20}, \zeta_6)$. The generators $\sqrt{2}$ and $\sqrt[6]{20}$ certainly lie in $K$, and since $$\zeta_6 \sqrt[6]{20}\cdot(\sqrt[6]{20})^5=20\zeta_6 \in K,$$ we conclude that $K=\mathbb Q( \sqrt{2},\sqrt[6]{20}, \zeta_6)$. A quick computation shows that $\zeta_6=\frac{1}{2}+i\frac{\sqrt{3}}{2}$ (or this is one option), so the splitting field is $\mathbb Q(\sqrt{2},\sqrt[6]{20},i\sqrt{3})$ as the OP claims.
