Suppose ladies are l1,l2 & l3.
Gents - g1,g2,g3.
any two and only 2 ladies sit together, so select 2 out of 3 ladies = 3C2 ways = 3
(lets say we selected l1,l2)
Now as only 2 of them can sit together, they should be surrounded by 2 gents. Select 2 gents out of 3 - 3C2 = 3 ways. (g1 and g2)
arrange these 4 now - possible ways - g1,l1,l2,g2 -b1
g2,l1,l2,g1 -b2
g1,l2,l1,g2 -b3
g2,l1,l2,g1 -b4
thus 4 ways or arranging these 4 people.
Consider one of the 4 selections as a box. (b1)
arrange g3 and l3 as - g3,b1,l3 or l3,b1,g3 (2 ways)
Thus total of 3*3*4*2 = 72 ways

IMO B
Alternative way to look at this problem is not to consider any restrictions and subtract the arrangements that do not satisfy the conditions.

Circular permutation of 6 people is (6-1)! = 5! = 120 ways
There are two cases that do not satisfy the conditions.
1) when all the ladies sit together, automatically gents will sit together
(L1 L2 L3 ) (G1 G2 G3) ---> (3! )(3!) =36


2) when ladies and gents sit alternatively.
--L1 G1 L2 G2 L3 G3---
Because this is a circular permutation ,we need to fix a position of one lady ,say L2 . Now other two ladies can be arranged in 2 ways(L1 and L3 can switch places) and gents can be arranged in 3! ways. Total arrangements = 2 * 3! = 2*6 = 12

SO # of ways any two and only two of the ladies sit together = 120 - (36+12) = 120-48 =72

Great explanations, guys!

One other way to look at this one is just by gauging the answer choices. If I know that, for the "two women and only two women must sit together" portion I'll have:

3 combinations (AB, AC, and BC)
multiplied by 2 ways that each can sit (AB or AC)

And that I'll have to multiply that portion by the number of overall arrangements for the whole group (in which the two women together comprises one of a number of items to be arranged since they'll most as one unit)...

Then I know that my answer must be a multiple of 6. It's 3*2*something. And only 72 fits the bill. Now, obviously there's that "none of these" option there so it's not foolproof, but one thing that's really nice about permutations/combinations problems is that often you can determine a couple essential factors of the correct answer and give yourself a pretty quick and educated 50/50 guess if not a direct process-of-elimination shot at the right answer without having to do all the work.