A uniform cube of mass m and side a is placed on frictionless horizontal surface. A vertical force $F=\frac{mg}{4}$ is applied to the edge as shown in figure. Normal reaction effectively acts at a distance.

Here,

The cube won't topple as there is no sufficient friction nor it will slide as there is no force in the horizontal direction.
For Translational equilibrium,

N + F = mg

$\therefore N=\frac{3mg}{4},\dots \left(i\right)$

Now,
Net torque about centre of the cube should be zero so that the cube doesn't rotate or N doesn't become 0.

$\begin{array}{l}{\left({\tau }_{\text{net }}\right)}_{CoM}=0\\ \Rightarrow \overline{F}\left(\frac{a}{2}\right)=Nx\dots \left(ii\right)\end{array}$

Using equation (i) and (ii),

$\begin{array}{l}x=\frac{a}{6}\\ \therefore y=\frac{a}{2}-x=\frac{a}{2}-\frac{a}{6}=\frac{a}{3}\text{ unit }\end{array}$