The coefficient of x 5 in the expansion of 2 x 3 - 1 3 x 2 5 is

General term of the binomial expansion of ${\left(2x^3-\frac{1}{3x^2}\right)}^5$ is

$T_{r+1}=5_{C_r}(2x^3{)}^{5-r} {\left(\frac{-1}{3x^2}\right)}^r$

$\Rightarrow T_{r+1}={}^{5}C_r{\left(2\right)}^{5-r}{\left(-\frac{1}{3}\right)}^r{x}^{15-5r}$

For coefficient of $x^5$, we must have

$15-5r=5\Rightarrow r=2$

So, required coefficient is

$$\begin{gathered} ={}^{5}C_2(2{)}^3\left(\frac{1}{9}\right) \\ =10\times 8\times \frac{1}{9} \\ =\frac{80}{9} \end{gathered}$$