Let $O$ be the center of $ABCD$'s circumsphere. Then the projections of $O$ on all the faces must be the circumcenters of these triangles.

Let $N$ be $O$'s projection on the face $ABD$. Because $\triangle{ABD}$ is equilateral, $N$ must be its center. Let $M$ and $P$ be midpoints of $CD$ and $AB$, respectively. Then $OM\perp CD$ and $N$ locates on $DP$. Let $\angle CDP=\alpha$. From the given conditions, we can derive: $$DP = \frac{3\sqrt{3}}{2}, CP=\frac{\sqrt{19}}{2}\implies \cos\alpha=\frac{\sqrt{3}}{3}, \sin\alpha=\frac{\sqrt{6}}{3}$$ Now consider $\triangle{DMN}$: $$DM =\frac{CD}{2} = 1, DN=\frac{2}{3}DP = \sqrt{3}$$ Hence, by the Law of Cosines: $$MN = \sqrt{1^2 +(\sqrt{3})^2 - 2\times 1\times \sqrt{3}\times\frac{\sqrt{3}}{3}} = \sqrt{2}$$ Finally, consider quadrilateral $ODMN$. Clearly $OD$ is the desired radius of $ABCD$'s circumsphere. Because $\angle{M}=\angle{N} = 90^\circ$, these four points are concyclic and $OD$ is its circumcircle's diameter. Applying the Law of Sine on $\triangle{DMN}$ yields: $$OD=\frac{MN}{\sin\alpha}=\boxed{\sqrt{3}}$$