We have the three circles such that they pass through the centers of the other two.

Now there are 4 sides of the square and only three circles, hence by pigeonhole principle, atleast two sides are tangent to the same circle. Clearly as the square subscribes the circles, the two sides should be adjacent.

So we can draw the figure as shown. AC the diagonal of the square will pass through the center P.

APEM is a square, ∴ AM = PM = PE = radius = 1

Similarly RNFD is a rectangle, hence DN = RF = 1

MN = PR cos(180 − ∠EPR)

PQR is clearly an equilateral triangle with side = 1

= −cos165 = cos15 = sin75

Hence, option 4.