= Solution
All roots are $\pm\sqrt2$ and $\pm\sqrt3$, so the <splitting field> is $K=\mathbb Q(\sqrt2,\sqrt3)$. We have $[\mathbb Q(\sqrt2):\mathbb Q]=2$.
To show that adjoining $\sqrt3$ doubles the degree, suppose $\sqrt3=a+b\sqrt2$ with rational $a,b$. Squaring and using irrationality of $\sqrt2$ gives $ab=0$. If $b=0$, then $a^2=3$; if $a=0$, then $b^2=3/2$. Neither is a rational square, as the prime exponents in a rational square are even. Thus $\sqrt3\notin\mathbb Q(\sqrt2)$. Its quadratic <polynomial> remains irreducible over that field, and the tower law gives
$$
\boxed{K=\mathbb Q(\sqrt2,\sqrt3),\qquad[K:\mathbb Q]=4.}
$$
A <basis> is $1,\sqrt2,\sqrt3,\sqrt6$.
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