RSA-FDH and random oracle query count
It is proportional to $poly(k)$. Here $k$ is the security parameter (or more strictly speaking it should be $(1^k)$) and intuitively the adversary's can only be allowed to query polynomial times (since it is a PPT algorithm) bounded in terms of $k$. Then $e$ and $e'$ only differs by a polynomial factor, and thus allows you to reduce what you want to prove to the RSA problem. 

The subtle technicality is that when modeling the adversary as a Turing machine and reasoning complexity (which is based on the size of input), the integer $k$'s length is only $log_2 k$, which will cause some formality problem. The more rigorous way is to supply $1^k$, a bit string of length $k$ that contains all 1s.