I'm going to assume that, judging by the fact that the coefficient of friction and angle of the incline seem to be intentionally left out, the point A refers to a point at the bottom of the incline, and the point B is located 19 m away from A on a horizontal surface. If this is not the case, please be sure to correct me!
Let v denote the speed of the box at point A, x the distance the box slides (without friction) down the incline, and θ the angle of the incline.
Recall that
v ² - v₀² = 2 a ∆x
where
v = final speed
v₀ = initial speed
a = acceleration
∆x = displacement
While on the incline, the box starts at rest (v₀ = 0) and slides a distance x from the starting point to point A with acceleration a to attain some speed v.
Use the free body diagram to determine the acceleration. By Newton's second law, the net force on the object acting parallel to the incline
∑ F = m g sin(θ) = m a → a = g sin(θ)
where m = 9.0 kg and g = 9.8 m/s².
Write the distance x in terms of θ with some trigonometry:
sin(θ) = (5.0 m) / x → x = (5.0 m) / sin(θ)
Now plug a and x into the formula above to solve for v :
v ² = 2 g sin(θ) ((5.0 m) / sin(θ))
v ² = (10.0 m) g
v ≈ 9.9 m/s
