Question 1101923
Here is the drawing of the cross section of the rods bundle to go with ikleyn's solution:
{{{drawing(400,400,-2.5,2.5,-2.3,2.7,
circle(-1,-0.5774,1),circle(1,-0.5774,1),
green(triangle(-1,-0.5774,1,-0.5774,0,1.1547)),
blue(line(-1,-0.5774,-1.866,-0.0744)),
blue(line(1,-0.5774,1.866,-0.0744)),
blue(line(-1,-0.5774,-1,-1.5574)),
blue(line(1,-0.5774,1,-1.5574)),
blue(line(0,1.1547,-0.866,1.6547)),
blue(line(0,1.1547,0.866,1.6547)),
circle(0,1.1547,1),red(line(1.866,-0.0744,0.866,1.6547)),
red(line(-1.866,-0.0744,-0.866,1.6547)),red(line(-1,-1.5574,1,-1.5574))
)}}}
The green triangle connects the centers of the 
The red line segments represent the portions of the band not in contact with the circles.
The end of those red segments are at {{{3cm/2=1.5cm}}} of the center of one of the circles,
That distance between (red) line and circle center is measured along the blue lines,
perpendicular to the red lines, of course.
That makes the green=blue-red quadrilateral a rectangle:
two congruent blue opposite sides make a parallelogram,
and a parallelogram with two right angles is a rectangle.
So, the green and te red segments are all conguent, all 3 cm long.
Of the {{{360^o}}} of central angle (and circumference) around each circle,
{{{60^o+90^o+90^o=240^o}}} is covered by
one angle of the triangle,
one angle of a rectangle, and
one angle of another rectangle.
Along the remaining {{{360^o-240^o=120^o}}} degrees of its circumference,
each circle is in contact with the bandwrapped around the rod bundle.