If the coefficients of $x^7$ and $x^8$ in ${\left(2+\frac{x}{3}\right)}^n$are equal, then value of $n$ is 

$T_{r+1},$ the $(r+1)th$ term in the expansion of  ${\left(2+\frac{x}{3}\right)}^n$is given by
                 $T_{r+1}{=}^n{C}_r{2}^{n-r}{\left(\frac{x}{3}\right)}^r{=}^n{C}_r\left(\frac{2^{n-r}}{3^r}\right)x^r$
  According to the given condition
$\begin{array}{l} { }^n{C}_7\left(\frac{2^{n-7}}{3^7}\right){=}^n{C}_8\left(\frac{2^{n-8}}{3^8}\right)\Rightarrow \frac{{ }^n{C}_7}{{ }^n{C}_8}=\frac{2^{n-8}}{3^8}\cdot \frac{3^7}{2^{n-7}}\\ \Rightarrow \frac{n!}{7!(n-7)!}\frac{8!(n-8)!}{n!}=\frac{1}{6}\Rightarrow \frac{8}{n-7}=\frac{1}{6}\\ \Rightarrow 48=n-7 \Rightarrow n=55\end{array}$