We have ${T_{r + 1}} = {\,^6}{C_r}{\left( {\frac{3}{2}{x^2}} \right)^{6 – r}}\left( { – \frac{1}{{3x}}} \right)$

$ = {\,^6}{C_r}{\left( {\frac{3}{2}} \right)^{6 – r}}{\left( {{x^2}} \right)^{6 – r}}{( – 1)^r}{\left( {\frac{1}{x}} \right)^r}\left( {\frac{1}{{{3^r}}}} \right)$

$ = {( – 1)^r}{\quad ^6}{C_r}\quad \frac{{{{(3)}^{6 – 2r}}}}{{{{(2)}^{6 – r}}}}\quad {x^{12 – 3r}}$

The term will be independent of $x$ if the index of $x$ is zero, i.e., $12-3 r=0 .$ Thus, $r=4$

Hence $5^{\text {th }}$ term is independent of $x$ and is given by ${( – 1)^4}{\,^6}{C_4}\frac{{{{(3)}^{6 – 8}}}}{{{{(2)}^{6 – 4}}}} = \frac{5}{{12}}$