Answer:
he sphere that uses less time is sphere A
Explanation:
Let's start with ball A, for this let's use the kinematics relations
         v² = v₀² - 2g (y-y₀)
indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0
          v² = 0 - 2 g (0- y₀)
          v =  
          v =  
          v = 4.427 √H
Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy
starting point. At the highest point, before you start to move
          Em₀ = U = m g y
final point. At the bottom of the ramp
          Em_f = K = ½ m v² + ½ I w²
notice that we include the kinetic energy of translation and rotation
energy is conserved
           Em₀ = Em_f
           mg H = ½ m v² + ½ I w²
angular and linear velocity are related
           v = w r
           w = v / r
the momentorot of inertia indicates that it is worth
           I =  m r²
 m r²
we substitute
            m g H = ½ m v² + ½ ( m r²) (
  m r²) ( )²
 )²
            gH =  v² +
  v² +  v² =
  v² =  v²
  v²
            v =  
            v =  
 
            v=3.742 √H
Taking the final speeds of the sphere, let's analyze the distance traveled, sphere A falls into the air, so the distance traveled is H.  The ball B rolls in a plane, so the distance (L) traveled can be found with trigonometry
            sin θ = H / L
            L = H /sin θ
we can see that L> H
In summary, ball A arrives with more speed and travels a shorter distance, therefore it must use a shorter time
Consequently the sphere that uses less time is sphere A