(note..I started typing this response before I saw Briman's answer..Thx again Briman)

Well this has been bugging me for some time. I thought someone here might know the honest answer off the top of their head. I thank Wing and Paul for giving it a shot. I decided I to try to figure it out for myself since no one really seems sure.

Here is what I found so far. ( I many not be correct, please feel to correct me if you know I am factual wrong.).

Wing...you are right about the twist rate and length relationship. All else being equal, a longer bullet requires a faster twist. However, the 12 and 20 ga shells are the same length. Therefore one would assume the projectiles are the same length. This does not seemas if it wouldexplain the large difference.

Paul...you are correct about velocity. All else being equal higher velocity requires a faster twist rate. The problem is, these barrels are designed for modern sabot shooting. Most modern 12 ga and 20 ga sabot shells (of the same design/brand) leave the barrel at very similar velocities. So that does not explain the very large difference.
Weight, although it is a factor in twist rates it is a relatively small one. In fact 12 ga and 20garifled barrels are designed to shot projectiles of such a varying degree of weights, that sometimes the weight of a 20 ga slug could weigh the same as a 12 ga slug of a different brand.

This leaves one factor I failed to consider thoroughly......Of course I was aware that the diameter of the projectile was a factor. But I didn't understand how it affected the relative relationship of a more important factor.

Here is the answer......Relative Length of the Projectile

It is the relationship of the diameter, (in inches),to the bullet length to diameter ratio, (length divided by diameter).

So...although both12 ga and 20 ga barrlesshot approximately the same length slugs,the20ga does so out of a barrel of less diameter. Therefore the Relative length of the projectile is greater.

This concept is known as the GREENHILL FORMULA....

Greenhill made easy:

T=(150 x D)/R
here: T is the twist required (number of inches for one revolu- tion),
D is the bullet diameter (in inches)
R is the bullet length to diameter ratio, (length divided by diameter)
Conversely, to find out what length bullet will be stabilized in a given twist, use:

L=(150 x D x D)/T

(that is, 150 x D squared divided by T)

where L= bullet length
The number 150 is a constant used by Greenhill and works well at velocities in the vicinity of 1500 fps or greater. At 2800 fps the constant can be changed to 180 with good results.