What is the coefficient of $a^4b^2$ in the expansion of $\left(2a-\frac{b}{3}\right)^6$?  Express your answer as a common fraction.
By the Binomial theorem, this term is: $$\binom64 (2a)^4\left(-\frac{b}{3}\right)^2=15\cdot16\cdot\frac{1}{9}a^4b^2=\boxed{\frac{80}{3}}a^4b^2$$