2012 AMC 10B Problem 16

Connect the centers of the three circles. This forms an equilateral triangle of side $4$, with area $4\sqrt3$. The included part of each circle is a $300^\circ$ sector, whose area is $\dfrac{300}{360}\pi\cdot2^2=\dfrac{10\pi}{3}$. The total area is $3\cdot\dfrac{10\pi}{3}+4\sqrt3=10\pi+4\sqrt3$. Thus, A is the correct answer.