jmc

We can multiply both sides of the congruence $6^{-1}\equiv 6^2(\mathrm{mod}m)$ by $6$: $$\underset{1}{\underbrace{6\cdot 6^{-1}}}\equiv \underset{6^3}{\underbrace{6\cdot 6^2}}(\mathrm{mod}m).$$Thus $6^3-1=215$ is a multiple of $m$. We know that $m$ has two digits. The only two-digit positive divisor of $215$ is $43$, so $m=\boxed{43}$.