Question 1061891
{{{drawing(300,300,-3.3,3.3,-3.3,3.3,
circle(0,0,3),
circle(-2,0,1),circle(2,0,1),
circle(1,1.732,1),circle(-1,1.732,1),
circle(1,-1.732,1),circle(-1,-1.732,1)
)}}} + another small circle = {{{drawing(300,300,-3.3,3.3,-3.3,3.3,
circle(0,0,3),green(circle(0,0,1)),
circle(-2,0,1),circle(2,0,1),
circle(1,1.732,1),circle(-1,1.732,1),
circle(1,-1.732,1),circle(-1,-1.732,1)
)}}}
The smaller circles' diameter is {{{1/3}}} of the diameter of the smaller circle. 
Radius of the larger circle = {{{18}}}
Radius of each smaller circle = {{{18/3=6}}}
Area of each smaller circle = {{{S=pi*6^2=36pi}}}
Since the larger circle is {{{3}}} times as wide,
its area, {{{L}}} , is {{{3^2=9}}} times larger:
{{{L=9S}}} ,
The area between the large circle and the six smaller circles is
{{{L-6S=9S-6S=3S=3(36pi)=highlight(108pi)}}} .
 
NOTE:
You may be familiar with this arrangement of 6 circles around a central one,
because you can form it with 7 pennies (or seven coins, or right cylindrical objects of the same shape).
Do you also see it in "pilings od same sized cylindrical objects,
such as a flatbed truck cargo of metal pipes or plastic tubes.
Do you want a geometry proof?
Joining the centers of adjacent circles with line segments will give you a regular hexagon.
Adding 3 diameters of the outside circle passing through the hexagon vertices,
you have  the hexagon split into 6 triangles.
The central angles measure {{{360^o/6=360^o}}} ,
and the triangle sides flanking those angles are congruent,
meaning that the triangles are isosceles, with a {{{60^o}}} vertex angle.
So, in the triangles, the base angles (adjacent to the hexagon sides)
must be congruent, and must measure
{{{(180^o-60^o)/2=120^o/2=60^o}}} .
So, those isosceles triangles (and all isoceles triangles with {{{60^o}}} vertex angles) are equilateral triangles.
All sides of those triangles are congruent,
all measuring 2 times the radius of the smaller circles.
That means that there is definitely just enough room in the center for another smaller circle.