The base-$ b $ representation of $ 2013 $ ends in $ 3 $ if and only if $ 2013 $ leaves a remainder of $ 3 $ when divided by $ b $: that is, if $ 2010 $ is a multiple of $ b $. Since $ 2010 = 2^1 \cdot 3^1 \cdot 5^1 \cdot 67^1, $ it has $ (1+1)(1+1)(1+1)(1+1) = 16 $ positive divisors. However, since $ 3 $ is a valid digit in base $ b, $ we must have $ b > 3, $ so we must subtract $ 3 $ from our count (since $ 1, $ $ 2, $ and $ 3 $ are all divisors of $ 2010 $). Therefore, the answer is $ 16 - 3 = \boxed{13} $.