jmc

The 10 in the numerator and the 55 in the denominator have a common factor of 5. Similarly, the $a^3$ and $a^2$ have a common factor of $a^2$. So we get $$\frac{10a^3}{55a^2}=\frac{2\cdot 5\cdot a^2\cdot a^1}{11\cdot 5\cdot a^2}=\frac{2\cdot \cancel{5}\cdot \cancel{a^2}\cdot a^1}{11\cdot \cancel{5}\cdot \cancel{a^2}}=\frac{2a}{11}.$$Substituting $a=3$ gives $\boxed{\frac{6}{11}}$.