There would be two forces acting on the box parallel to the floor, with a net force of
∑ F = p - f = m a
where p = magnitude of the push, f = mag. of friction, m = mass of the box, and a = acceleration. To find p, we first need f .
There are also only two forces acting on the box perpendicular to the floor, with net force
∑ F = n - w = 0
where n = mag. of normal force of the floor on the box and w = weight of the box. The net force is 0 because the box is only accelerating parallel to the floor.
w = m g, where g = 9.8 m/s² is the mag. of the acceleration due to gravity, so we can solve for n :
n = w = m g
n = (22 kg) (9.8 m/s²)
n = 215.6 N
The kinetic friction is proportional to the normal force by a factor of the given coefficient of friction, µ = 0.17, such that
f = µ n
f = 0.17 (215.6 N)
f = 36.652 N
Now solve for the required pushing force:
p - 36.652 N = (22 kg) (1.9 m/s²)
p ≈ 78 N