2019 AMC 12A Problem 10

Place a unit circle at the center, six around it with centers at distance $2$ (a hexagon), and six more with centers at distance $2\sqrt{3}$ in the outer gaps. That is $1 + 6 + 6 = 13$ circles. The outermost circles are tangent to the big circle, whose radius is therefore $2\sqrt{3} + 1.$ Its area is $$\pi(2\sqrt{3} + 1)^2 = \pi(13 + 4\sqrt{3}).$$ Subtracting the $13$ unit circles leaves $\pi(13 + 4\sqrt{3}) - 13\pi = 4\pi\sqrt{3}.$ Thus, the correct answer is A.