Let $\omega$ be the angular velocity of the cube (just after the bullet strikes) about an axis passing through D. 

Conservation of angular momentum about this axis gives

 $mv\left(\frac{2\ell }{3}\right)=I\omega$  where I is the moment of inertia about the axis through D. 

I =  $\left[\frac{M{\ell }^2}{6}+\frac{M{\ell }^2}{2}\right]=\frac{2M{\ell }^2}{3}$ 

 $\Rightarrow \omega =\left(\frac{mv}{M\mathcal{l}}\right)$  

The cube will topple if the center of mass is just able to rise from  $\left(\frac{l}{2}\right)\text{ to }\left(\frac{l}{\sqrt{2}}\right)$. In such a case, the rotational energy must be equated to the change of potential energy. 

Thus, $\frac{1}{2}I{\omega }^2=Mg\left(\frac{\ell }{\sqrt{2}}-\frac{\ell }{2}\right)$  

Using the values of I and $\omega$, we get the expression for v that will just topple the cube: 

 $v=\frac{M}{m}{\left[3gl\left(\frac{\sqrt{2}-1}{2}\right)\right]}^{\frac{1}{2}}$.