Take the volume of the room (measured)
Find the volume of a ping-pong ball (measured)
Find the void fractional volume of the spherical shaped-balls (estimated)
Void fractional volume is the amount of "wasted space" between adjacent ping-pong balls
NOTE: there are several programs on how to estimate this value.
If I were to estimate the fractional void volume without relying on any computer program and just wanted to make a good engineering approximation, I would approximate the sphere to a cube with similar dimensions and add some "fudge factor"
Number of ping pong balls = Volume of room / (Volume of single ping-pong ball)
EXAMPLE:
Given a room 3.5 m x 3.5 m x 2. m = 24.5 m^3 (as an example)
Regulation ping pong ball has a diameter of 40 mm
The volume of a single ping pong ball is (4/3)(pi)r^3 = 3.351 x 10^-5 m^3
The volume of a cube with similar dimension (each edge measures 40 mm) = 6.4 x 10^-5 m^3
Ratio of volume of sphere to cube = 52.36%
Number of ping pong balls = (24.5 m^3)/(6.4 x 10^-5) = 382,813 ping pong balls (assuming cube-shape)
Add a conservative "fudge factor" of approximately 30% ====>: approximately 497,657 ping-pong balls
I would say conservative because I could justify a "fudge factor" of up to 52.36%; but I am taking into account that the balls will not pack perfectly. (FYI: some sources recommend adding up to 40% as a "fudge factor").
This would be my best "guess" although it may be a bit low. I would be comfortable in approximating this to maybe three significant figures at most, meaning the answer I would give would be about 498,000 ping-pong balls