Answer
$a.\quad R=3. \quad $Interval of convergence:$\quad -3\leq x\leq 3$
$b.\quad $ Interval of absolute convergence:$\quad -3\leq x\leq 3$
$c.\quad $ No other values of x for which the series converges conditionally.
Work Step by Step
(See text: "How to Test a Power Series for Convergence".)
$\text{Step 1.}$
Use the Ratio Test to find the interval where the series converges absolutely.
$\begin{align*}
\displaystyle \frac{u_{n+1}}{u_{n}}&=\displaystyle \frac{x^{n+1}}{(n+1)\sqrt{n+1}\cdot 3^{n+1}}\div\frac{x^{n}}{n\sqrt{n}\cdot 3^{n+1}}\\
&=\displaystyle \frac{x^{n+1}}{(n+1)\sqrt{n+1}3^{n+1}}\cdot\frac{n\sqrt{n}3^{n}}{x^{n}}\\
&=\displaystyle \frac{x}{3}\cdot\frac{n}{n+1}\cdot\sqrt{\frac{n}{n+1}}\\
\displaystyle \lim_{n\rightarrow\infty}|\frac{x}{3}\cdot\frac{n}{n+1}\cdot\sqrt{\frac{n}{n+1}}|&\lt 1\displaystyle \\
|\displaystyle \frac{x}{3}|\lim_{n\rightarrow\infty}\frac{n}{n+1}\cdot\lim_{n\rightarrow\infty}\sqrt{\frac{n}{n+1}}&\lt 1\displaystyle \\
|\displaystyle \frac{x}{3}|(1)\cdot\sqrt{\lim_{n\rightarrow\infty}\frac{n}{n+1}}&\lt 1\displaystyle \\
|\displaystyle \frac{x}{3}|&\lt 1 \displaystyle \\
-1&\lt\displaystyle \frac{x}{3} \lt 1 \\
-3&\lt x\lt 3 \end{align*}$
Determine the center and radius:
$a-R\lt x\lt a+R$
$0-3\lt x\lt 0+3$
$a=0,\quad $
the radius is $R=3,$
the interval of absolute convergence is $-3\lt x\lt 3$
$\text{Step 2.}$
If the interval of absolute convergence is finite, test for convergence or divergence at each endpoint.
$ x=-3\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}\frac{(-3)^{n}}{n\sqrt{n}\cdot 3^{n}}=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n\sqrt{n}}\qquad$
... $|\displaystyle \frac{(-1)^{n}}{n\sqrt{n}}|\rightarrow 0,\qquad $... series is absolutely convergent
$ x=3\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}\frac{3^{n}}{n\sqrt{n}\cdot 3^{n}}=\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}}=\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}\qquad$
... a convergent p-series $(p=\displaystyle \frac{3}{2}\gt 1).$
Absolute convergence interval is $\qquad -3\leq x\leq 3$
$\text{Step 3.}$
If the interval of absolute convergence is $a-R\lt x\lt a+R$,
the series diverges for $|x-a|\gt R.$
So,
$a.\quad R=3. \quad $Interval of convergence:$\quad -3\leq x\leq 3$
$b.\quad $ Interval of absolute convergence:$\quad -3\leq x\leq 3$
$c.\quad $ No other values of x for which the series converges conditionally.
