When the sixth ball was dropped, 5 seconds have elapsed
 from the time the first ball was dropped, 4 seconds after the
 second ball was dropped, 3 seconds after the the third ball was 
 dropped, 2 seconds after the fourth ball was dropped and 1 second 
 after the fifth ball was dropped.

 Thus being said, when the sixth ball was dropped & since the 3rd ball
 has been falling for 3 seconds, then

 D3 = (1/2)(9.8)(3^2) = 44.1 meters

 Since the fourth ball has been falling for 2 seconds,

 D4 = (1/2)(9.8)(2^2) = 19.6 meters

 and since the fifth ball has been falling for 1 second,

 D5 = (1/2)(9.8)(1)^2 = 4.9 m
 
						

							
When the sixth ball was dropped,the initial velocity will be zero.Considering the time taken by the 3rd,4th and the 5th ball as 1/3,1/2 and 1,we can easily find the height of 3rd ball (H1) from the top of building ,height of 4th ball (H2) and height of 5th ball (H3).
 
Here, u=0, a=(9.8) m/s
 
By,
    S=ut+1/2at^2,                   {ut is eliminated because u =0}
  1. -H1 = –½ x (9.8) x 9
       (H1)= ½ x 88.2
             =(44.1)meters
 
  2. -H2 = – ½ x (9.8) x 2
      (H2)= ½ x 19.6
            =(9.8)meters
 
 3. -H3 = – ½ x (9.8) x 1
     (H3)= ½ x 9.8
           =(4.9)meters
Hence,when the 6th ball is being dropped,the position of the 3rd,4th,5th balls from the top of the building are 44.1m,19.6m and 4.9m respectively.