The $(r+1)$ th term in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$ is given by

${T_{r + 1}} = {\,^{10}}{C_r}{\left( {\sqrt {\frac{x}{3}} } \right)^{10 – r}}{\left( {\frac{3}{{2{x^2}}}} \right)^r}$

$ = {\,^{10}}{C_r}\frac{{{x^{5 – \left( {r/2} \right)}}}}{{{3^{5 – \left( {r/2} \right)}}}} \cdot \frac{{{3^r}}}{{{2^r}{x^{2r}}}}$

$ = {\,^{10}}{C_r}\frac{{{3^{(3r/2) – 5}}}}{{{2^r}}}{x^{5 – (5r/2)}}$

For $T_{r+1}$ to be independent of $x,$ we must have $5-(5 r / 2)=0$ or $r=2 .$ Thus, the ${3^{rd}}$ term is independent of $x$ and its coefficient is given by

$^{10}{C_2}\frac{{{3^{3 – 5}}}}{{{2^2}}} = \frac{{10 \times 9}}{2} \times \frac{{{3^{ – 2}}}}{4} = \frac{5}{4}.$