The free body diagram for the given scenario is shown below-

From the equilibrium of the vertical component of forces, it can be written that
N====mg−Fsin30∘mg−2F100−2F2200−F........................(1)
From the equilibrium of horizontal component of forces, it can be written that,
Fcos30∘=μN......................(2)
Substitute the expression for the normal reaction force from equation (1) into equation (2) and solve to calculate the value of the applied force.
32F43FF===≈0.25×(2200−F)200−F43+120025.2
