Solution

\section{Kinematics}
Let us consider the instantaneous kinematics of the system.\
Let $M$ be the mass of the wedge.\
Let $V$ be the velocity of the wedge. \
Let $v$ be the velocity of the block relative to the wedge.
\vspace{6pt} \
The velocity components of the block relative to the ground are:
$$v_x = v\cos{\alpha}-|V|, \qquad v_y= v\sin{\alpha}$$

From there we obtain \begin{equation} \tan{\beta}=\frac{v\sin{\alpha}}{v\cos{\alpha}-|V|} \end{equation}
\section{Displacement of CM}
Since no force acts on the system horizontally, the horizontal displacement of the center of mass $\Delta x_{cm_x}$ must equal to 0, therefore we can write:

$$\Delta x_{cm_x} = \frac{m \vec{\Delta x}_{mx}+M \vec{\Delta x}_{Mx}}{m +M}$$ \vspace{1pt}
$$0 = \frac{mv_x\Delta t - M|V| \Delta t}{m+M}$$
\vspace{1pt}
\begin{equation} M|V|=m(v \cos{\alpha}-|V|) \end{equation}
\vspace{1pt}
$$|V|=\frac{mv\cos{\alpha}}{M+m}$$
\section{Substitution}
Now we substitute V into (1):
$$\tan{\beta}=\frac{v}{v-\frac{mv}{M+m}}\tan{\alpha}$$
And obtain the following: \boldmath $$M=\frac{m\tan{\alpha}}{\tan{\beta} + \tan{\alpha}}$$