The given equation of circle is
$x^2+y^2-2 x+k y+4=0$ ...(i)
Centre of above circle (i) is
$c \equiv\left(1,-\frac{k}{2}\right)$
$\because$ Circle $S$ is concentric with circle (i)
$\therefore \quad$ Equation of $S$ is
$(x-1)^2+\left(y+\frac{k}{2}\right)^2=r^2$ ...(ii)
The centre $\left(1, \frac{-k}{2}\right)$ lies on the line $3 x-2 y+4=0$
$\therefore \quad 3 \times 1-2 \times\left(\frac{-k}{2}\right)+4=0 \Rightarrow k=-7$
Putting the value of $k$ in eqn. (ii), we get
$(x-1)^2+\left(y-\frac{7}{2}\right)^2=r^2$ ...(iii)
$\because$ Eqn. (iii) passes through the point $(3,-2)$. Then, we have
$\begin{aligned} & (3-1)^2+\left(-2-\frac{7}{2}\right)^2=r^2 \\ & \Rightarrow r^2=4+\frac{121}{4}=\frac{137}{4} \\ & \Rightarrow r=\frac{\sqrt{137}}{2}\end{aligned}$
$\therefore$ Radius of the circle $S$ is, $r=\frac{1}{2} \sqrt{137}$.