6. In the tetrahedron $ABCD$, it is known that
$$
\begin{array}{l}
\angle ADB=\angle BDC=\angle CDA=60^{\circ}, \\
AD=BD=3, CD=2 .
\end{array}
$$

Then the radius of the circumscribed sphere of the tetrahedron $ABCD$ is
6. $\sqrt{3}$.

As shown in Figure 3, let the circumcenter of tetrahedron $ABCD$ be $O$. Then point $O$ lies on the perpendicular line through the circumcenter $N$ of $\triangle ABD$ and perpendicular to the plane $ABD$.
By the given conditions, $N$ is the center of the equilateral $\triangle ABD$.
Let $P$ and $M$ be the midpoints of $AB$ and $CD$ respectively. Then point $N$ lies on $DP$, and $ON \perp DP$, $OM \perp CD$.
Let the angle between $CD$ and the plane $ABD$ be $\theta$. Then $\cos \theta = \frac{1}{\sqrt{3}}$, $\sin \theta = \frac{\sqrt{2}}{\sqrt{3}}$.
In $\triangle DMN$,
$DM = \frac{1}{2} CD = 1$,
$DN = \frac{2}{3} DP = \frac{2}{3} \times \frac{\sqrt{3}}{2} \times 3 = \sqrt{3}$.
By the cosine rule,
$MN^2 = 1^2 + (\sqrt{3})^2 - 2 \times 1 \times \sqrt{3} \times \frac{1}{\sqrt{3}} = 2$.
It is easy to see that $MN = \sqrt{2}$.
Therefore, the radius $R$ of the sphere $O$ is $OD = \frac{MN}{\sin \theta} = \sqrt{3}$.