Solution:
$\omega_{A}$, $\omega_{B}$, $\omega_{C}$ intersect at the circumcenter; thus, every point within the circumcircle, and no point outside of it, is in two or more circles. The area inside exactly two circles is shaded in the figure. The two intersection points of $\omega_{A}$ and $\omega_{B}$, together with $A$, form the vertices of an equilateral triangle. As shown, this equilateral triangle cuts off a "lip" of $\omega$ (bounded by a $60^{\circ}$ arc of $\omega$ and the corresponding chord) and another, congruent lip of $\omega_{B}$ that is not part of the region of interest. By rotating the first lip to the position of the second, we can reassemble the equilateral triangle. Doing this for each of the 6 such triangles, we see that the desired area equals the area of a regular hexagon inscribed in $\omega$. The side length of this hexagon is $(2 / 3) \cdot (\sqrt{3} / 2) \cdot 2 = 2 \sqrt{3} / 3$, so its area is $6 \cdot (\sqrt{3} / 4) \cdot (2 \sqrt{3} / 3)^{2} = 2 \sqrt{3}$, and this is the answer.