yup, holly's answer pretty much nailed it on the head. but just to go into more detail:
basic premise:
understand which of the compounds "want" to lose their proton. the compound that "wants" to lose its proton the most will have the most acidic proton.
with the simple alkanes, they usually never want to lose their protons because they're stable and happy w/all sigma bonds and 4 bonds around the c atoms. the only cases where they'd be deprotonated would be w/a very strong, reactive metal such as Li or Na. so scratch these out.
benzene, as holly said...you just don't want to do anything to alter its stable state. it's very happy as a conjugated aromatic system and if you removed a proton, you'd mess up the aromaticity. so scratch out benzene.
so the answer has to be between the two cyclodienes. with dienes, the only proton in question is going to be the allylic proton (the one on the c atom next to the dbl bond). looking at 1,3cyclohexadiene: draw a picture. now deprotonate the C-5 hydrogen. you can see that the compound is definitely stable as the electrons can be delocalized on the adjacent carbons. however, the compound isn't aromatic, b/c all the carbons aren't sp2 hybridized (c-6 is sp3 hybridized).
now compare this to removing an allylic proton from the c-5 of cylclopentadiene. now you actually have an aromatic (read: very very stable) system b/c all the carbons are planar/sp2. so 1,3 cyclopentadiene without a proton is very stable, happy.
thus of all the compounds 1,3 cyclopentadiene is "happiest" without a proton and thus this is going to be the compound that most wants to give up its proton....and from our basic premise, carries the most acidic proton. 🙂
