Consider the below diagram
Moment of the force F about point A, ${\mathrm{\tau }}_1$ = $\mathrm{rF}$ (anti-clockwise)
Moment of weight mg of the cube about point A.

${\mathrm{\tau }}_2=\mathrm{mg}\times \frac{\mathrm{a}}{2}\left(\text{ clockwise }\right)$

Step 2: Equate the torques for no motion and find the value of F.

$\text{ Cube will not exhibit motion, if }{\mathrm{\tau }}_1={\mathrm{\tau }}_2$

($\because$ In this case, both the torque will cancel the effect of each other)

$\therefore \mathrm{F}\times \mathrm{a}=\mathrm{mg}\times \frac{\mathrm{a}}{2}\Rightarrow \mathrm{F}=\frac{\mathrm{mg}}{2}$

Step 3: Find the value of F for which cube will rotate.

Cube will rotate only when, ${\tau }_1>{\tau }_2$
$\Rightarrow \mathrm{F}\times \mathrm{a}>\mathrm{mg}\times \frac{\mathrm{a}}{2}\Rightarrow \mathrm{F}>\frac{\mathrm{mg}}{2}$

Step 4: If the normal reaction is acting at a/3 from point A and for no motion find F and interpret.
Let the normal reaction is acting at a/3 from point A, then
$\mathrm{mg}\times \frac{\mathrm{a}}{3}=\mathrm{F}\times \mathrm{a}\text{ or }\mathrm{F}=\frac{\mathrm{mg}}{3}\text{ (For no motion) }$

When $\mathrm{F}=\frac{\mathrm{mg}}{4}$ which is less than mg/3.                          $(\mathrm{F}<\frac{\mathrm{mg}}{3})$
there will be no motion,

$\begin{array}{llll}\text{∴ (a) }\to \text{ (ii) }& \text{ (b) }\to \text{ (iii) }& \text{ (c) }\to \text{ (i) }& \text{ (d) }\to \text{ (iv) }\end{array}$