A cubical block of side ‘L’ rests on a rough horizontal surface with coefficient of friction $\mathrm{\mu }$ . A horizontal force F is applied on the block as shown in figure. The minimum force required to topple the block is

Normal response N will pass through point P at the critical state. In this scenario, when ${\tau }_N=0={\tau }_{fr}$ (about P), the block will collapse when 

${\mathrm{\tau }}_{\mathrm{F}}>{\mathrm{\tau }}_{\mathrm{mg}} \mathrm{or}$

$\mathrm{FL}>\left(\mathrm{mg}\right)\frac{\mathrm{L}}{2}$

$\therefore \mathrm{F}>\frac{\mathrm{mg}}{2}$

Therefore,$\mathrm{F}=\frac{\mathrm{mg}}{2}$ is the minimal force needed to overturn the block.
​Hence the correct answer is $\mathrm{F}=\frac{\mathrm{mg}}{2}$.