What Happens To The Cube?

From the figure above, we see that the normal reaction on the incline is the perpendicular component of the cube's weight, i.e. $Mg \cos{A}$ and the force that pulls it down the plane is the other component, i.e. $Mg \sin{A}$ .

Since, the coefficient of friction $\mu$ is greater than $\tan {A}$ , and so, by multiplying $Mg \cos{A}$ on both the sides, we can write:

$\mu Mg \cos{A} \ge Mg \sin{A}$

which implies that the object won't slide down the plane.

Now, for the remaining two options, if we try and find the clockwise torque about the poinr of contact, O , we can write, ${\tau}_{CW} = \dfrac {a} {2} Mg \sin{A}$ , and the anticlockwise torque can be written as ${\tau}_{ACW} = \dfrac {a} {2} Mg \cos{A}$ , where $a$ is the edge length of the cube.

Since $A$ is greater than ${45}^{\circ}$ , so, $\sin{A} \ge \cos{A}$ , and as a result ${\tau}_{CW} \ge {\tau}_{ACW}$ , which causes the cube to topple about this point.