6. In $\triangle A B C$, $\angle B A C=90^{\circ}, A C=\sqrt{3}$, $A B=4$, $D$ is a point on side $B C$, $\angle C A D=30^{\circ}$. Then the length of $A D$ is ( ).
(A) $\frac{6}{5}$
(B) $\frac{7}{5}$
(C) $\frac{8}{5}$
(D) $\frac{9}{5}$
6. C.

As shown in Figure 7, draw $D E \perp A C$, with the foot of the perpendicular at $E$. Let $E D=x$, then $A D=2 x$, $A E=\sqrt{3} x$, and $C E=\sqrt{3}-\sqrt{3} x$.
From $\triangle C D E \backsim \triangle C B A$, we get
$$
C E: C A=D E: B A \text {, }
$$

which is $(\sqrt{3}-\sqrt{3} x): \sqrt{3}=x: 4$.
Therefore, $x=\frac{4}{5}$.
Thus, $A D=2 x=\frac{8}{5}$.