I don't think this is a perfectly valid question. Your result can be pretty much exact within your method, and at the same time far from the best you can get in DFT - and even farther from what is experimentally relevant.

First of all, take note that the accuracy of your result (given the particular geometry of your molecule and the method used - your basis set, your XC functional etc.) will depend on the thresholds you set in your SCF calculation/geometry optimization. With finite basis sets, you never reach the actual energy, and using a larger set will not always get you closer to it. You can use different thresholds, a different basis set and a different XC functional and look at how much your energy changes - then you'll see how much it depends on the particular computational setup. That's why many QM papers are concerned with benchmarkng, and you can see for yourself in the literature that the results will vary wildly for different approaches.

Secondly, it usually makes little sense to talk about absolute energies. Calculate your interaction energy, convert it to a more intuitive unit such as kcal/mol; while absolute energies will vary a lot between e.g. XC functionals, for relative quantities the setup-specific systematic error cancels to a large extent. Anyway, be prepared for differences up to several kcal/mol for different setups. Personally, I would round up the result in kcal/mol to a single decimal place (and that may answer your question - for practical purposes, people usually do just that).

And there's also geometry. If you do your calculations for just one lowest-energy configuration, you'll effectively model your system frozen at 0 Kelvins. This particular configuration may contribute a lot in a real-life system at finite temperature, but thermal fluctuations will add some variability, and thus the 'actual' (average) interaction energy will again differ from what you obtain in your calculations.

Thanks Milosz, It looks like the answer is complicated...I was using intuitively one decimal place; now I am more sure