80 should be the answer, you don't need brute force for this... think burnside+frobenius+cauchy
EDIT: I mistakenly place 157 there.. It's 80
Here's how you do it:
You consider the spots aroud the circle as distinct, and look at how it behaves under rotations
Burnside's lemma states that the number of orbits in a permutation group is equal to the average number of fixed points under each permutation...
applied here, it would mean that:
The number o different placements(orbits) is equal to the average number of placements which does not change(these are fixed points) under different amounts of rotation(there are 12 possibilities)
For rotation of 0 degrees, every placement is a fixed point: there are 12C = 924 of them
For rotation of 30 and 330 degrees there are no configurations which can be fixed
for rotation of 60 and 300 degrees, there are 2 configurations which are fixed for each (4 total) (both alternating white and black, one starting with white, the otehr starting with black)
and the others:
90,270: none
120:240: 4C2 = 6 each
150,210 : none
180: 6C3 = 20
And then just apply the lemma:
number of distinct placements = (1/12)*(924 + 2 + 6 + 20 + 6 + 2) = 80
(I said 157 first time coz I made a mistake and added another 924 there :S)