Given: Mass of bullet $m$, the velocity of bullet $v_0$ and mass of the wooden block $M$.

Let the velocity of the bullet after the collision with the wooden block be $v$

Applying the law of conservation of momentum, we obtain

$m{v}_0=\left(M+m\right)v$

$v=\frac{m{v}_0}{\left(M+m\right)}$

Now applying the law of conservation of mechanical energy, we obtain

$\frac{1}{2}m{v_0}^2=mgh$

Putting the value of $v$ in the above equation, we get  

$\frac{1}{2}\left(M+m\right){\left(\frac{m{v}_0}{\left(M+m\right)}\right)}^2=\left(M+m\right)gh$

Maximum height of the block-bullet system $h=\frac{1}{2}m{v}_0^2\left(\frac{M}{m+M}\right)$