jmc

Since $\gcd (m,n)=12$, both $m$ and $n$ are divisible by 12. Then $10m$ is divisible by $10\cdot 12=120$, and $15n$ is divisible by $12\cdot 15=180$. Since 60 divides both 120 and 180, $\gcd (10m,15n)$ must be at least 60.

If we set $m=n=12$, then $\gcd (m,n)=\gcd (12,12)=12$, and $\gcd (10m,15n)=\gcd (120,180)=60$, which shows that the value of 60 is attainable. Therefore, the smallest possible value of $\gcd (10m,15n)$ is $\boxed{60}$.