6. C.

Connect $M C, M N$, and draw $M E \perp A C$ at point $E$.
$$
\begin{array}{l}
\text { Then } \angle A M E=\frac{1}{2} \angle A M C=\angle A B D=60^{\circ} \\
\Rightarrow \frac{A M}{A C}=\frac{A M}{2 A E}=\frac{1}{\sqrt{3}} .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
\triangle A M N \cong \triangle B M N, \triangle A M N \backsim \triangle A C D . \\
\text { Therefore } \frac{S_{\text {quadrilateral } A M B N}}{S_{\triangle A C D}}=\frac{2 S_{\triangle A M N}}{S_{\triangle A C D}}=2\left(\frac{A M}{A C}\right)^{2}=\frac{2}{3} .
\end{array}
$$
6. As shown in Figure $3, \odot M$ and 
$\odot N$ intersect at points $A$ and $B$, a line $CD$ through point $B$ intersects $\odot M$ and $\odot N$ at points $C$ and $D$ respectively, and $\angle ABD = 60^{\circ}$. Then the ratio of the area of quadrilateral $AMBN$ to the area of $\triangle ACD$ is ( ).
(A) $1: 2$
(B) $1: 3$
(C) $2: 3$
(D) $3: 4$