The equation of the circle passing through $(2,0),(0,1) and (4,5) $ is
$3\left(x^{2}+y^{2}\right)-13 x-17 y+14=0$
This passes through $(0, c)$.
$\therefore 3 c^{2}-17 c+14=0$
$\Rightarrow c=1, \frac{14}{3}$
Since, $c=1$ is already there, for point (0,1)
Therefore, we take $c=\frac{14}{3}$.