Given,
$x^{2}+y^{2}-2 x-4 y+c=0\,\,\,...(i) $
and $ x^{2}+y^{2}-4 x-2 y+4=0\,\,\,...(ii)$
Centre of first circle is $(1,2)$ and radius of this circle is $\sqrt{1+4-c}=\sqrt{5-c}$
Centre of second circle is $(2,1)$ and radius of this circle is$\sqrt{4+1-4}=\sqrt{1}=1$
Hence, $C_{1} \equiv(1,2), r_{1}=\sqrt{5-c}$
and $C_{2}=(2,1), r_{2}=1$
Angle between the two circles is
$\cos \theta=\frac{\left(C_{1} C_{2}\right)^{2}-\left(r_{1}^{2}+r_{2}^{2}\right)}{2 r_{1} r_{2}}$
$\cos \theta=\frac{\left[(2-1)^{2}+(1-2)^{2}\right]-(5-c+1)}{2 \sqrt{5-c}(1)}$
$\Rightarrow \cos 60^{\circ}=\frac{1+1-5+c-1}{2 \sqrt{5-c}}$
$\Rightarrow \frac{1}{2}=\frac{c-4}{2 \sqrt{5-c}}$
$\Rightarrow \sqrt{5-c}=c-4$
On squaring both sides, we get
$5-c =c^{2}-8 c+16 $
$\Rightarrow c^{2}-7 c+11 =0$
$\therefore c =\frac{7 \pm \sqrt{49-44}}{2}$
$=\frac{7 \pm \sqrt{5}}{2}$