Let m be the mass of the puck. Then the upward normal force exerted by the ice has magnitude n = mg (since Newton's law says the net force on the puck in the vertical direction is ∑ F = n - mg = 0).
Then the frictional force applies a force of magnitude f = 0.052mg, and Newton's second law says the net horizontal force acting on the puck is
∑ F = -f = ma
Solve for the acceleration a :
-0.052mg = ma
a = -0.052 (9.8 m/s²)
a ≈ -0.510 m/s²
Assuming constant friction, the puck slides to a rest over a distance x such that
0² - (5.30 m/s)² = 2ax
Solve for x :
x = -(5.30 m/s)²/(2a)
x = -(5.30 m/s)² / (2 (-0.510 m/s²))
x ≈ 27.6 m
