6. In tetrahedron $ABCD$, it is known that $\angle ADB=\angle BDC=\angle CDA=60^{\circ}$, $AD=BD=3$, $CD=2$. Then the radius of the circumscribed sphere of tetrahedron $ABCD$ is $\qquad$.
As shown in the figure, construct a regular tetrahedron $A B C^{\prime} D$ with edge length 3. Let the center of the base $A B D$ be $O$, and connect $D O$ to intersect $A B$ at point $E$. Construct $C^{\prime} H \perp D O$ at point $H$. The center of the circumscribed sphere of the tetrahedron $A B C D$ lies on the line $C^{\prime} O$. Since
$$
\begin{array}{l}
D E=\frac{3 \sqrt{3}}{2} \Rightarrow O D=\sqrt{3} \Rightarrow O H=\frac{\sqrt{3}}{3}, \\
C^{\prime} O=\sqrt{6} \Rightarrow C H=\frac{2 \sqrt{6}}{3}
\end{array}
$$
$\Rightarrow O C=\sqrt{3}=O D$. Therefore, $O$ is the center of the circumscribed sphere of the tetrahedron $A B C D$, and the radius is $\sqrt{3}$.