Vertically, the object is in equilibrium, so that the net force in this direction is
∑ F (vertical) = n - mg = 0
where n is the magnitude of the normal force due to the contact between the object and surface. You're given that the object's weight is mg = 550 N, so n = 550 N as well.
Horizontally, the net force would be
∑ F (horizontal) = p - f = 0
where p is the magnitude of the applied force and f is the magnitude of (kinetic) friction opposing p. Now,
f = 0.012n = 0.012 (550 N) = 6.6 N
so that you need to apply a force of p = 6.6 N to keep the object sliding at a steady pace.