6. In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=30^{\circ}, A C=2, M$ is the midpoint of $A B$, fold $\triangle A C M$ along $C M$, so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. At this moment, the volume of the tetrahedron $A-B C M$ is equal to
$$
\frac{2 \sqrt{2}}{3}
$$
6.【Analysis and Solution】The tetrahedron $A-BCM$ after folding is shown in the figure. Take the midpoint $D$ of $CM$, and connect $AD$. In $\triangle BCM$, draw $DE \perp CM$ intersecting $BC$ at $E$, and connect $AE$, then
$$
\begin{array}{l}
AD=\sqrt{3}, DE=CD \cdot \tan 30^{\circ}=\frac{\sqrt{3}}{3}, \\
CE=2 DE=\frac{2 \sqrt{2}}{3} .
\end{array}
$$

In $\triangle ABC$, $AC=2, AB=2 \sqrt{2}, BC=2 \sqrt{3}$.
So $AC^{2}+AB^{2}=BC^{2}$.
Therefore, $\angle BAC=90^{\circ}$.
In $\triangle ACE$, $AE^{2}=AC^{2}-CE^{2}-2 AC \cdot CE \cos \angle ACE=\frac{8}{3}$.
So $AE^{2}+CE^{2}=4=AC^{2}$.
Also, $AE^{2}+DE^{2}=3=AD^{2}$, thus, $AE \perp BC, AE \perp DE$. Therefore, $AE \perp$ plane $BCM$.
So $V_{A-BCM}=\frac{1}{3} AE \cdot S_{\triangle BCM}=\frac{2 \sqrt{2}}{3}$.