6. B.

As shown in Figure 7, connect $BD$, $AC$, extend $BA$ and $CD$, and let them intersect at point $P$.

In the right triangle $\triangle ACP$, it is easy to prove that $D$ is the midpoint of $PC$,
$$
\begin{array}{l}
PD = CD = AD \\
= 2 \sqrt{3}.
\end{array}
$$

Notice that,
$$
\text{Rt } \triangle BCD \cong \text{Rt } \triangle BPD.
$$

Thus, $PB = BC$.
By the secant theorem,
$$
\begin{array}{l}
PA \cdot PB = PD \cdot PC \\
\Rightarrow (BC - 2) BC = 2 \sqrt{3} \times 4 \\
\Rightarrow BC = 6.
\end{array}
$$

Therefore, the area of $\odot O$ is $9\pi$.
6. As shown in Figure 2, quadrilateral $ABCD$ is an inscribed quadrilateral in circle $\odot O$ with diameter $BC$, and $D$ is the midpoint of arc $\overparen{AC}$. If $AB=2$, $AD=2\sqrt{3}$, then the area of $\odot O$ is ( ).
(A) $8\pi$
(B) $9\pi$
(C) $10\pi$
(D) $12\pi$