Given situation is

Forces acting on the block are

In above diagram,

$ma=$ pseudo force due to acceleration of plank, $f=$ force of static friction, $mg=$ weight of block and $N=$ normal reaction.

Here, we must note that due to motion of plank, normal reaction force does not passes through centre of mass.

If line of action of normal force at a distance $x$ from centre, then for equilibrium of block, net torque about centre of mass must be zero.

$\Rightarrow {\tau }_{\mathrm{Normal} \mathrm{reaction}}={\tau }_{\mathrm{Friction}}$

$mg$ and $ma$ does not produces any torque about centre of mass as their line of action passes through centre of mass.

$\Rightarrow N·x=f·l/2$

$\Rightarrow mgx=ma\frac{l}{2}$

$\left[\because N=mg \& f=ma\right]$

$x=\left(\frac{a}{g}\right)\frac{l}{2}$

or $x=\frac{1}{10}·\frac{l}{2} \left[\because a=\frac{g}{10}\right]$

or $x=\frac{l}{20}$

So, coordinates of point from which reaction passes are $\left(-\frac{l}{20},0\right)$