Okay, so let's see what we've got here. You're asked to predict the reaction rates of some ketones. They are all equal in structure with the exception of IV, which has no ketone group. So you would think that IV would be the fastest to react, but there's some special tricks up compound I's sleeve.
In compound I, you have a hydroxyl group one carbon over from the ketone. Ketones are stabilized by alkyl groups on either side, which this compound has (two methyls on the C=O bond). Since this is the most stable ketone, it will react the fastest, even faster than the alcohol. You might wonder why III doesn't react equally, since the ketone has two alkyl groups to stabilize it as well. But remember that little hydroxyl group over there, he's going to want to steal some of the glory away from the ketone. When it is in the same orientation as the ketone (as in compound I), the ketone will be protonated faster, since it is more electron withdrawing than the hydroxyl group. In compound III, it is on the end of the molecule, so there is no real competition going on. II is obviously the least stable since the hydroxyl group is an electron withdrawing group which destabilizes the carbonyl group.
In conclusion, I is the most stable ketone and therefore the most reactive. II is the least stable ketone and therefore the least reactive. III is a stable ketone, but the stereochemistry of the compound lowers its reaction rate. IV is an alcohol and as previously mentioned by the original poster, acid-base reactions proceed very quickly due to carbocation rearrangement. I'm not sure of the exact numbers, but I'm pretty sure that a stable ketone will react faster than a secondary carbocation.
So that's my reasoning as to the given answer. I feel like there's more to it with carbocation rearrangement and resonance stabilization, but hopefully this will at least drive you in the right direction. I probably didn't explain something in there right either, so anybody please feel free to correct me.