What is the force required to produce a 1.4 nm torque when applied to a door at a 60.0º angle and 0.40 m from the hinge?

Refer to the diagram shown below.

The component of the applied force perpendicular to the door is
F * sin(60°) = 0.866F N

Because the moment arm is 0.40 m, the torque is
(0.866F N)*(0.4 m) = 0.3464F N-m

This torque is equal to 1.4 N-m, therefore
0.3464F = 1.4
F = 4.04 N

4.04 N

Ithink it is letter b, i may be wrong but i think that is the answer