Three blocks of 1N 2N and 3N are placed one over the other respectively. If the friction between the topmost 1N and the middle 2N block is twice the friction between the lowest 3N and floor then determine the minimum horizontal force to disturb the equilibrium. (I tried to solve, the photo contains my Solution.)
Your answer is correct. But I don't think acceleration needs to be involved for this case, rather it can be solved by statics as depicted in the figure below. From the figure, it is clear that in order to mobilize the middle block, the force must be equal to the larger friction force on the top and bottom faces, $F \geq 6\mu N$.