jmc

The prime factorization of 240 is $2^4\cdot 3\cdot 5$. We want $c$ to be as small as possible, so $c=2$. Now we have $a\cdot b=2^3\cdot 3\cdot 5=120$. For the maximum $b$, we seek the minimum $a$ while $b<a$. If $b<a$ then $a\cdot b<a^2$, so $120<a^2$. That means $a$ is at least 11. However, 11 is not a factor of 240. The least factor of 240 greater than 11 is $2^2\cdot 3=12$. So the minimum $a$ is 12 and that makes the maximum $b=\frac{2^3\cdot 3\cdot 5}{2^2\cdot 3}=2\cdot 5=\boxed{10}$.