rotation of a rigid body

Frictionless wall means,
 the only force exerted by the wall on the ladder is normal to the wall; say F.outward.

 

The force of friction exerted by the ground on the ladder is along the goround, horizontally inward. These two must be equal for translational equiibrium. F = f

 

Also, for rotational euilibrium, taking momentum about the point of contact with the ground,

Anticlockwise momentum is of the force F at distance L = 3.0 m at angle of 60^o.

Moment of this force, FxL cos θ = FL x0.866. ---(1)

 

Clockwise moment of the weight mg of the ladder acting in the middle.

This is  mg.(L/2)sin 30 = mgLx0.5/2 = mgL/4  __(2)

From (1) and (2)

                  FL x0.866 = mgL/4 

But F = f = μ.mg;

Substituting

              μ.mg.L = mgL/(4x0.866)

                 μ = 1/(4x0.866) = 0.2886 = 0.289