jmc

First, we find the prime factorization of $6300$ to be $2^2\cdot 3^2\cdot 5^2\cdot 7$. Note that the odd divisors of 6300 are precisely the integers of the form $3^a{5}^b{7}^c$ where $0\le a\le 2$, $0\le b\le 2$, and $0\le c\le 1$. Note also that distributing $(1+3+9)(1+5+25)(1+7)$ yields 18 terms, with each integer of the form $3^a{5}^b{7}^c$ (again, where $0\le a\le 2$, $0\le b\le 2$, and $0\le c\le 1$) appearing exactly once. It follows that the sum of the odd divisors of 6300 is $(1+3+9)(1+5+25)(1+7)=\boxed{3224}$.