The numbers 4725 and 126, written as the products of their prime factors, are 4725=3^3 x 5^2 x 7. The smallest positive integer value of m for which 126m, is a multiple of 4725.
To find the smallest positive integer value of $m$ such that $126m$ is a multiple of $4725$, we need to find the prime factorization of $4725$. As given, $4725=3^3 \cdot 5^2 \cdot 7$.
So, to make $126m$ a multiple of $4725$, we need to determine which prime factors of $4725$ are missing from $126$ and multiply $m$ by these primes.
The missing prime factors are: $3$, $5$, and $7$.
Therefore, $m$ must be multiplied by $3 \cdot 5 \cdot 7 = \boxed{105}$ to make $126m$ a multiple of $4725$.