3. Given that $B D$ is the median of $\triangle A B C$, $A C=6$, and $\angle A D B=45^{\circ}, \angle C=30^{\circ}$. Then $A B=(\quad)$.
(A) $\sqrt{6}$
(B) $2 \sqrt{3}$
(C) $3 \sqrt{2}$
(D) 6
3. C.

As shown in Figure 3, draw a perpendicular from point $B$ to $AC$, intersecting the extension of $CA$ at point $H$. Then
$$
\begin{array}{l}
H D=H B, H C=\sqrt{3} H B \\
\Rightarrow H C-H D=(\sqrt{3}-1) H B=3 \\
\Rightarrow H B=\frac{3}{2}(\sqrt{3}+1), H A=\frac{3}{2}(\sqrt{3}-1) \\
\Rightarrow A B=\sqrt{H A^{2}+H B^{2}}=3 \sqrt{2} .
\end{array}
$$