3. Given that $AB$ is the diameter of $\odot O$, $C$ is a point on $\odot O$, $\angle CAB=15^{\circ}$, and the angle bisector of $\angle ACB$ intersects $\odot O$ at point $D$. If $CD=\sqrt{3}$, then $AB=(\quad)$.
(A) 2
(B) $\sqrt{6}$
(C) $2 \sqrt{2}$
(D) 3
3. A.

As shown in Figure 3, connect $O C$, and draw $O M \perp C D$ at point $M$.
Then $\angle O C M$
$$
=45^{\circ}-15^{\circ}=30^{\circ} \text {. }
$$

Thus, $C M=\frac{\sqrt{3}}{2} O C$.
Therefore, $A B=2 O C$
$$
=\frac{4}{\sqrt{3}} C M=\frac{2}{\sqrt{3}} C D=2 \text {. }
$$