In the expansion ${\left(2+\frac{1}{3x}\right)}^n$

$(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$ be the general term of expansion  ${(x+a)}^n)$

Then $T_{r+1}={}^n{C}_r{2}^{n-r}{\left(\frac{1}{3x}\right)}^r={}^n{C}_r\frac{2^{n-r}}{3^r}{x}^{-r}$

So, coefficient of $x^{-r}={}^n{C}_r\frac{2^{n-r}}{3^r}$  

$\Rightarrow$  Coefficient of $x^{-7}={}^n{C}_7\frac{2^{n-7}}{3^7}$

And   coefficient of $x^{-8}$  $={}^n{C}_8\frac{2^{n-8}}{3^8}$  

According to problem coefficients are equal hence

$\begin{array}{c}\frac{{}^n{C}_7{2}^{n-7}}{3^7}=\frac{{}^n{C}_8{2}^{n-8}}{3^8}\\ \Rightarrow \frac{{}^n{C}_8}{{}^n{C}_7}=\frac{3^8{2}^{n-7}}{3^7{2}^{n-8}}\\ \Rightarrow \frac{{}^n{C}_8}{{}^n{C}_7}=6\end{array}$

$\Rightarrow \frac{{}^n{C}_8}{{}^n{C}_7}=6\Rightarrow \frac{n-7}{8}=6$  $\begin{array}{c}(\therefore \frac{{}^n{\text{C}}_r}{{}^n{\text{C}}_{r-1}}=\frac{n-r+1}{r})\\ \end{array}$

$\therefore n=55$