Let us consider the given diagram.
Moment of the force due to $F$ about point is anticlockwise $A, \tau_1=\vec{F} \times a$
Moment of force due to weight $m g$ of the cube about point $A$ is clockwise
$\tau_2=m g \times \frac{a}{2}$


(i) Cube will not move or exhibit motion, if $\tau_1=\tau_2$
( $\because$ In this case, both the torque will cancel the effect of each other)
$$
\therefore \vec{F} \times a=m g \times \frac{a}{2} \Rightarrow F=\frac{m g}{2} \text { so (a) } \rightarrow \text { (ii) }
$$
(ii) Cube will rotate if, $\tau_1>\tau_2$
$$
\vec{F} \times a>m g \times \frac{a}{2} \Rightarrow F>\frac{m g}{2} \text { so (b) } \rightarrow \text { (iii) }
$$
(iii) If normal reaction is effectively acting at $\frac{a}{3}$ from point $A$, then
Torque due to $m g, \tau_2=m g \times \frac{a}{3}$
Torque due to $F$ is, $\tau_1=F \times a$
(For no motion)
So, $\tau_1=\tau_2 \Rightarrow F \cdot a=m g\left(\frac{a}{3}\right)$
If $F>m g$, block move upside, (c) $\rightarrow$ (i)
(iv) If $F=\frac{m g}{4}$ which is less than $\frac{m g}{3}$, or $\left(\frac{m g}{3}\right)>\left(\frac{m g}{4}\right)$ So due to $F=\frac{m g}{4}$, block will not move. (d) $\rightarrow$ (iv) there will be no motion.
$\therefore$ (a) $\rightarrow$ (ii), (b) $\rightarrow$ (iii), (c) $\rightarrow$ (i), (d) $\rightarrow$ (iv)