This is not easy to answer in a general way, as the rate of a photochemical reaction depends on a lot of things, only one of which is the light intensity. What is sure is that light must be absorbed in order to make a photochemical reaction go, but the relationship is not only linear. In fact, the rate of absorption of photons isn't even linear, and it can be more or less linear depending on the molecule that's doing the absorbing. If your molecule has a large absorption cross-section and is already efficiently photoactivated at the lower light intensity, doubling the intensity isn't going to do much because your rate is limited by other factors. If, however, your cross-section is small and your light absorption efficiency is low (and your quantum yield of reaction is fast), then your rate limiting step is absorption, and doubling your intensity will certainly make a substantial difference. You've also got other kinetic things to worry about because, unlike a traditional reaction, your reagents aren't already necessarily premixed. This kind of inner filter effect essentially limits the rate of mixing of your reagents (one of which is light). This is why vigorous stirring MAY be important.... it also depends on how clear your solution is! So, physical theory isn't necessarily going to give you the practical answer you're looking for.
If you tell a little bit more about the reaction you're doing, this will help. If you prefer to look into it on your own, I suggest finding a copy of Turro's excellent book Modern Molecular Photochemistry, which has a whole chapter on rates and also sections on homolytic cleavage reactions. Otherwise, if you don't really care about learning all you're probably going to need to learn, I suggest the trial and error method: try the lower Watt bulb and if you're happy with the result, move on with your life.