I've narrowed it down to either $20$ or $40$:
$$x^{10}-5=0\iff x^{10}=5e^{2\pi ik}\iff x=5^{1/10}e^{\pi ik/5}, k=0,1,2,3,4$$
One can show that the splitting field is $\mathbb{Q}(5^{1/10},e^{\pi i/5})$, and the degree of this extension is divisible by both $[\mathbb{Q}(5^{1/10}:\mathbb{Q}]=10$ and $[\mathbb{Q}(e^{\pi i/5}:\mathbb{Q}]=4$.
However, I am unsure of whether the minimal polynomial of $e^{\pi i/5}$, $\Phi_{10}(x)= x^4-x^3+x^2-x+1$, factors (into quadratics) over $\mathbb{Q}(5^{1/10})$ or whether $x^{10}-5$ factors over $\mathbb{Q}(e^{\pi i/5})$. Equivalently, I am unsure of the degrees $[\mathbb{Q}(5^{1/10},e^{e\pi i/5}):\mathbb{Q}(e^{\pi i/5})]$ and $[\mathbb{Q}(5^{1/10},e^{e\pi i/5}):\mathbb{Q}(5^{1/10})].$
How can either of the latter two degrees be computed?
$\Bbb Q(\sqrt 5)$ is a subextension of both $\Bbb Q(5^\frac 1 {10})$ and $\Bbb Q(\exp(\frac {i \pi}5))$.
(letting $\zeta = \exp(\frac {i \pi}5))$, try to see what $(\zeta + \zeta^2 - \zeta^3 - \zeta^4)^2$ is)
Since they are extensions of $\Bbb Q(\sqrt 5)$ of degrees $5$ and $2$ respectively, and $5$ and $2$ are coprime, their composition $L = \Bbb Q(5^\frac 1 {10}, \exp(\frac {i\pi}5))$ is an extension of $\Bbb Q(\sqrt 5)$ of degree $10$, so $L$ is an extension of $\Bbb Q$ of degree $20$.