2020/21 AHSMC Part 1 Problem 16

Let $r$ be the radius of a smaller circle. The centres of the smaller circles are the vertices of an equilateral triangle of side $2r$. The height of this equilateral triangle is $r\sqrt{3}$. The centre of the larger circle is the same as the centre of the equilateral triangle, which is located twice as far from each vertex as from the opposite edge. Thus the radius of the larger circle is $r + \dfrac{2r\sqrt{3}}{3}$. Now, $$r + \frac{2r\sqrt{3}}{3} = 10 \iff r = \frac{30}{2\sqrt{3}+3} = 10(2\sqrt{3}-3).$$ The answer is (B).