Answer
The radius of convergence: $\frac{1}{3}$
The interval of convergence: $[-\frac{1}{3},\frac{1}{3}]$
Work Step by Step
Let $a_n=\frac{(-3)^nx^n}{n\sqrt{n}}$.
Find the interval such the series $\sum_{n=1}^\infty a_n$ is convergent:
$\lim |\frac{a_{n+1}}{a_n}|<1$
$\lim \left|\frac{\frac{(-3)^{n+1}x^{n+1}}{(n+1)\sqrt{n+1}}}{\frac{(-3)^nx^n}{n\sqrt{n}}}\right|<1$
$\lim |\frac{-3n\sqrt{n}x}{(n+1)\sqrt{n+1}}|<1$ (Note $\lim \frac{n\sqrt{n}}{(n+1)\sqrt{n+1}}=1$)
$|-3x|<1$
$|x|<\frac{1}{3}$
Check the convergence at the ends of interval:.
For $x=\frac{1}{3}$, the series would be $\sum_{n=1}^\infty (-1)^n\frac{1}{n\sqrt{n}}$ and it is convergent by the Alternating Series Test.
For $x=-\frac{1}{3}$, the series would be $\sum_{n=1}^\infty \frac{1}{n\sqrt{n}}$ and it is also convergent as the p-series with $p=1.5$.
Thus, the radius of convergence is $\frac{1}{3}$ and the interval of convergence is $[-\frac{1}{3},\frac{1}{3}]$.
