A block is placed on a smooth inclined plane at an angle ' $\theta$ ' to the horizontal. The acceleration must the plane be moved horizontally so that the block to fall freely is

Since the block is in contact with the wedge, the acceleration along the normal to the incline is zero. 

And from Newton’s 2nd law of motion we know that $F_{net}=m{a}_{net}$.
Therefore, the net force on the block in the direction of the normal to the incline is zero. And from the figure we get that the net force along the normal is $F_{net}=ma\sin \theta +N-mg\cos \theta$.
But $F_{net}=0$.
$\Rightarrow ma\sin \theta +N-mg\cos \theta =0$
$\Rightarrow ma\sin \theta +N=mg\cos \theta$ …. (i)
We have already discussed that the block will fall freely when the normal force is zero.
Therefore, put $N=0$ in (i).
$\Rightarrow ma\sin \theta +0=mg\cos \theta$
$\Rightarrow a\sin \theta =g\cos \theta$
$\therefore a=g{\displaystyle \frac{\cos \theta }{\sin \theta }}=g\cot \theta$.