3. As shown in Figure $2, A B$ is tangent to $\odot O$ at point $A$, connecting point $B$ with a point $D$ inside $\odot O$ intersects $\odot O$ at point $C, A B$ $=6, D C=C B=3$, $O D=2$. Then the radius of $\odot O$ is . $\qquad$
3. $\sqrt{22}$.

As shown in Figure 5, extend $B D$ to intersect $\odot O$ at point $E$, and extend $O D$ to intersect $\odot O$ at points $F$ and $G$, where $F G$ is the diameter of $\odot O$. Let the radius of $\odot O$ be $r$.
By the secant-tangent theorem, we have
$$
B C \cdot B E = B A^{2} \text {, }
$$

which means $3(D E + 6) = 6^{2} = 36$.
Therefore, $D E = 6$.
By the intersecting chords theorem, we get
$$
D E \cdot D C = D F \cdot D G \text {, }
$$

which means $6 \times 3 = (r - 2)(r + 2)$.
Therefore, $18 = r^{2} - 4 \Rightarrow r = \sqrt{22}$.