6. In tetrahedron $ABCD$, one edge has a length of 3, and the other five edges have a length of 2. Then the radius of its circumscribed sphere is
6. $\frac{\sqrt{21}}{3}$.

As shown in Figure 2, let
$$
\begin{array}{l}
B C=3, \\
A B=A C=A D \\
=B D=C D \\
=2,
\end{array}
$$
$E$ and $F$ are the midpoints of $B C$ and $A D$, respectively. The projection of point $D$ on plane $A B C$ is $H$, which is the circumcenter of $\triangle A B C$, and the center of the circumscribed sphere $O$ lies on $D H$.
Since $\triangle A B C \cong \triangle D B C$, we have $A E=D E$.
Thus, $E F$ is the perpendicular bisector of $A D$.
Therefore, the center of the sphere $O$ is the intersection of $D H$ and $E F$, and it is also the orthocenter of isosceles $\triangle E A D$.
Let the radius of the sphere be $r$.
From $\triangle D O F \backsim \triangle E A F$, we get $r=\frac{A E \cdot D F}{E F}$.
And $A E^{2}=2^{2}-\left(\frac{3}{2}\right)^{2}=\frac{7}{4}, D F=A F=1$,
$E F^{2}=A E^{2}-A F^{2}=\frac{3}{4}$,
Therefore, $r=\sqrt{\frac{7}{3}}=\frac{\sqrt{21}}{3}$.