If  $z=x+iy$  then the centre of the circle  $\left|\frac{z-3}{z-2i}\right|=2$  is

$\left|\frac{x+iy-3}{x+iy-2i}\right|=2\Rightarrow \frac{\left|\left(x-3\right)+iy\right|}{\left|x+i\left(y-2\right)\right|}=2$

$\Rightarrow \frac{\sqrt{{\left(x-3\right)}^2+y^2}}{\sqrt{x^2+{\left(y-2\right)}^2}}=2$

$\Rightarrow \frac{x^2+9-6x+y^2}{x^2+y^2+4-4y}=4$

$\Rightarrow x^2+y^2-6x+9=4x^2+4y^2-16y+16$

$\Rightarrow 3x^2+3y^2+6x-16y+7=0$

$\Rightarrow x^2+y^2+2x-\frac{16}{3}y+\frac{7}{3}=0$

$\Rightarrow$ centre = $\left(-1,\frac{8}{3}\right)$