Anonymous: If W plays at a, she has 2 eyes. Therefore, B only needs to consider 2 options. It looks like 'b' kills.
If W cuts off the 3 black stones, B can make them into Nakade. (Dieter: must be played at , because otherwise White plays there, not at like in this diagram.)
and at the square-marked stone. Saving the 3 black stones at 'a' and saving by playing at the circle-marked stone are miai for B to reduce W to only 1 eye. (Dieter: but after WHite a and b, White has miai in the corner for two eyes)
Dieter: the end result is ko. I agree this is an expert level problem. I didn't see before looking at the situation after , so you cannot say I have properly solved it.
unkx80: Almost there, you saw but missed . Actually for me the problem diagram was totally misprinted, and I had to reconstruct it from the solution diagram, so I never got a chance to solve it at all.
makes miai of a and the marked stone.