The polar coordinates equivalent to $(-3,\sqrt{3})$ are

$\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}=\sqrt{9+3}=2\sqrt{3}$

$\mathrm{\alpha }={\tan }^{-1}\frac{\left|\mathrm{y}\right|}{\left|\mathrm{x}\right|}$

$\mathrm{\alpha }={\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)$

$\mathrm{\alpha }={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\mathrm{\pi }}{6}$

Point $(-3,\sqrt{3})$ lies in the second quadrant. 

$\therefore \mathrm{\theta }=\mathrm{\pi }-\mathrm{\alpha }=\mathrm{\pi }-\frac{\mathrm{\pi }}{6}=\frac{5\mathrm{\pi }}{6}$

So, polar coordinates are $(2\sqrt{3},\frac{5\mathrm{\pi }}{6})$