This is challenging...

For $n$ overlapping circles, there are $n$ points where the circumferences of adjacent circles meet which make up a regular $n$ -side polygon. The two vertices of the polygon and the center point form an isosceles triangle inscribed in the circle. Let the arc extended by the side of the polygon to have a length of $a$ and that by the equal side be $b$ then the length of a visible arc of a circle is $l = a+b$ . We note that $a+2b = 1$ , the circumference of a circle. Also that arc $a$ extends an angle of $\frac {4\pi}n$ at the center of the circle. Then $a = \frac {4\pi}n \times \frac 1{2\pi} = \frac 2n$ and $b = \frac {1-a}2 = \frac 12 - \frac 1n$ , implying $l = a+b = \frac 12 + \frac 1n$ . Since there are $n$ visible arcs in total, the total length of visible arcs is $nl = \frac n2 + 1$ . When $\frac n2 + 1 = 60$ , $\implies n = \boxed{118}$ .

Each visible arc of a circle subtends an angle of $π+\dfrac{2π}{n}$ , so that the length of the arc is $\dfrac{1}{2}+$ $\dfrac{1}{n}$ . The total length of the visible arcs is therefore $1+\dfrac{n}{2}$ , which measures $60$ units. Therefore $n=118$

general formula

Circumference is given =1 (n+2)(pi)R=60

》{(n+2)2(pi)R}/2=60

》{(n+2)(1)}=120

》n=118👌

$\text{ Joining the center of each adjacent circles we get a } n\text{-sided regular polygon whose circumcenter lies at point }C \text{ , the common point of all the circles. Let the radius of each circle be } r$ .

$OC=OA=r=O'C=O'A\Rightarrow AOCO' \text{ is a rhombus. } \newline \Rightarrow AC\perp OO'$

$\large\angle OCO' = \frac{2\pi}{n}$

$\large\Rightarrow \angle BCO' = \frac{\pi}{n}$

$\large\Rightarrow \angle BO'C = \frac{\pi}{2} - \frac{\pi}{n} = \frac{(n-2)\pi}{2n}$

$\large\Rightarrow \angle AO'C=2\angle BO'C = \frac{(n-2)\pi}{n}$

$\text{Length of visible arc of circle with center } O' = \text{ Circumference of circle } - \text{ invisible length of arc of circle with center } O'\newline \hspace{190pt} = \large 1 - \frac{\angle AO'C}{2\pi}\cdot \small\text{ circumference of circle }\newline \text{ } \newline \large\hspace{190pt} = 1 - \frac{(n-2)\pi}{2n\pi}\cdot 1\newline\text{ }\newline\hspace{190pt} = \frac{n+2}{2n} \small\quad\dots Eq. \; 1$

$\text{Length of visible arcs is } 60$ units.

$\text{Length of visible arc of circle with center } O' = \frac{60}{n} \quad\dots Eq. \; 2\newline \text{ Using }Eq.\;1\text{ and }Eq.\;2\,, \text{ we get }\newline\large\frac{60}{n} = \frac{n+2}{2n}\newline\Rightarrow n+2=60\times 2 = 120\newline\Rightarrow n=\boxed{118}$