For A to remain stationary with respect to B, friction and it's weight should cancel out i.e. $\mu N = (8m)g$ ; where N is the normal force by B's surface to A and they should have the same acceleration.

N will have the magnitude but opposite direction of pseudo force due to it's acceleration.

N = (8m) $\ddot{x}$ .

Thus,

$\mu (8m)\ddot{x} = (8m)g$

$\ddot{x} = \frac{g}{\mu}$

Now the system of A and B is connected to C by an inextensible string. So for A and B to accelerate at $\ddot{x}$ C should also accelerate at $\ddot{x}$ .

The force for the whole system A,B and C to accelerate at $\ddot{x}$ is thw weight of C.

So the force equation is

$(8m + 2m + M)\ddot{x} = Mg$ ; $M$ is the mass of C.

$\ddot{x} = \frac{Mg}{10m + M}$

So,

$\frac{g}{\mu} = \frac{Mg}{10m + M}$

$M = \frac{10m}{\mu - 1}$ .

This is the minimal solution of M as increasing it will only cause to increase in the normal force N which increases limit of the friction then required.