 Solution

Let $\alpha =2^{1/6}=\sqrt{2}/\sqrt[3]{2}$. Then $L=\mathbb{Q}(\alpha ,{\zeta }_3)$ is the splitting field of $X^6-2$. Eisenstein gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=6$, and adjoining nonreal ${\zeta }_3$ doubles it, so the degree is 12 and the extension is Galois. Valuation at a prime above 2 forces $n\mid 6$ if $x^n=2$; conversely $x={\alpha }^{6/n}$ works for $n=1,2,3,6$.