Answer
$ R=2 $
$-2\lt x\lt 2$
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{2^{n}}{3 n}(x+3)^{n}$$
Since $a_n =\frac{2^{n}}{3 n}(x+3)^{n}$ and $a_{n+1} =\frac{2^{n+1}}{3 n+3}(x+3)^{n+1}$, then
\begin{aligned}
\rho &=\lim_{n\to \infty }\left|\frac{a_{n+1}}{a_{n}}\right|\\
&=\lim_{n\to \infty }\frac{|x|^{n+1}}{2^{n+1}} \cdot \frac{2^{n}}{|x|^{n}}\\
&=\frac{|x|}{2}
\end{aligned}
Then the series converges for
$$\frac{|x|}{2} \lt 1 \ \to -2\lt x\lt2 $$
The radius of convergence is $ R=2 $.
Now, we check the end points
For $x= -2 $ $$\sum_{n=1}^{\infty} \frac{2^{n}}{3 n}(x+3)^{n} =\sum_{n=1}^{\infty} (-1)^n $$
which is a divergent alternating series.
For $x=2 $ $$\sum_{n=1}^{\infty} \frac{2^{n}}{3 n}(x+3)^{n}= \sum_{n=1}^{\infty}(1)^n $$
which is divergent by the divergence test.
Hence, the interval of convergence is $$-2\lt x\lt 2$$
