If the angle between two equal circles with centres $(-2,0),(2,3)$ is ${120}^0$, then the radius of the circle is

The distance between the given two centers is $d=\sqrt{16+9}=5$

We know that, if $\theta$ is angle between two circles having radii $r_1,r_2$ and distance between centers is $d$ then $\cos \theta =\frac{d^2-r_1^2-r_2^2}{2r_1{r}_2}$

Given $r_1=r_2=r$ and $\theta ={120}^0$, $d=5$

Hence, 

$\begin{array}{rcl}-\frac{1}{2}& =& \frac{25-2r^2}{2r^2}\\ -2r^2& =& 50-4r^2\\ 2r^2& =& 50\\ r^2& =& 25\\ r& =& 5\end{array}$