Answer
$\Sigma_{n=0}^{\infty}\frac{(x-3)^{n}2^{n}e^{6}}{n!}$
$R=\infty$
Work Step by Step
$e^{2x}=\Sigma_{n=0}^{\infty}\frac{(x-3)^{n}2^{n}e^{6}}{n!}$
$\lim\limits_{n \to \infty}|\frac{\frac{(x-3)^{n+1}2^{n+1}}{(n+1)!}}{\frac{(x-3)^{n}2^{n}}{n!}}|$
$=|x-3|\lim\limits_{n \to\infty}|\frac{2}{n+1}|$
$=0\lt 1$
The the radius of convergence is $R=\infty$.