6. In $\triangle A B C$, it is known that $\angle C=90^{\circ}, \angle B=30^{\circ}$, $A C=2$, $M$ is the midpoint of $A B$, and $\triangle A C M$ is folded along $C M$ such that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Then the volume $V$ of the tetrahedron $A-B C M$ is ( ).
(A) $\frac{\sqrt{2}}{3}$
(B) $\frac{2}{3}$
(C) $\frac{\sqrt{6}}{3}$
(D) $\frac{2 \sqrt{2}}{3}$
6. D.

As shown in Figure 3, draw $B D \perp C M$, and let it intersect the extension of $C M$ at point $D$. Draw $A F \perp C M$ at point $F$, and draw $E F / / B D$ such that $E F = B D$. Connect $B E$.
Clearly, $A F = F E = B D = \sqrt{3}$, and $E B = D F = 2$.
Thus, $A E^{2} = A B^{2} - E B^{2} = 8 - 4 = 4$.
The height $h$ of the tetrahedron $A-B C M$, which is the distance from point $A$ to the plane $B C M$, is also the distance from point $A$ to side $E F$ in the isosceles $\triangle A E F$.
By the equality of areas, we have
$$
h=\frac{2 \times \sqrt{3-1}}{\sqrt{3}}=\frac{2 \sqrt{6}}{3} \text {. }
$$

Therefore, $V=\frac{1}{3} \times \frac{1}{2} \times 2 \sqrt{3} \times \frac{2 \sqrt{6}}{3}=\frac{2 \sqrt{2}}{3}$.