Clarifications:

1. The height of the room?

2. The room is occupied or furniture

3. Height of the room?

4. Only one layer of the floor is filled or the balls can be stacked upon each other?

Assumptions:

1. Capacity is utilized to the fullest

Only one layer and balls are not stacked upon each other:

- This case is fairly simple all you need to do is

1 feet = 30.48 cms

The diameter of the basketball ~25 cms

16 feets = 30.48 * 16 = 488 cms

The number of balls in one column = 488/16 = 19

The number of columns = 488/16 = 19

Some of the space remains unoccupied, as the entire space to fit a ball is not available

The answer = 30* 30 = 900;

Additional height required to stack another layer:

(refer the calculations in the attached image)

~21.65

Height of the room average ~ 10 feet = 304.8 cms

Number of layers in the stack possible =

First layer requires complete space 304.8 - 25 = 279.8

Number of subsequent layers = 254.8/21.65 = ~11.76

With 11 as the ball must occupy the space completely:

= 11 + 1 (bottom layer) = 13 layers and 361 balls in each layer

= 11* 361

= 3971

With 13 as the ball as the earlier unoccupied space would allow to bottom balls to spread and allows some space in height:

12 + 1 intital layer = 13 * 361

= 4693 balls

My recommendation - it will easily release enough height attributed to the Spread due to the incomplete space in 2D and 14 layers would be possible hence answer is 4693