A person sitting on the top of a tall building is dropping balls at...

Let the time when the 6th ball is dropped be $t = 0$.

Since the balls are dropped at regular intervals of $1 \text{ s}$, the time elapsed for the other balls are:
Time for 5th ball, $t_5 = 1 \text{ s}$
Time for 4th ball, $t_4 = 2 \text{ s}$
Time for 3rd ball, $t_3 = 3 \text{ s}$

The distance fallen by a freely falling body in time $t$ is given by $h = \dfrac{1}{2}gt^2$. Taking $g = 9.8 \text{ m/s}^2$:

Distance of the 3rd ball from the top:
$h_3 = \dfrac{1}{2} \times 9.8 \times (3)^2 = 4.9 \times 9 = 44.1 \text{ m}$

Distance of the 4th ball from the top:
$h_4 = \dfrac{1}{2} \times 9.8 \times (2)^2 = 4.9 \times 4 = 19.6 \text{ m}$

Distance of the 5th ball from the top:
$h_5 = \dfrac{1}{2} \times 9.8 \times (1)^2 = 4.9 \times 1 = 4.9 \text{ m}$

The positions of the 3rd, 4th, and 5th balls are $44.1 \text{ m}$, $19.6 \text{ m}$, and $4.9 \text{ m}$ below the top respectively.