This problem was solved by Dhara Fernando, 13, from the University High School in Victoria, Australia.

Let the radii of the two inner circles be $r$ and $R$, then the radius of the surrounding circle is $R+r$. You are given that the area inside the largest circle surrounding the two smaller circles inside it is equal to the area of the larger of the two internal circles, so

${\displaystyle \pi (R+r){}^2-\pi R^2-\pi r^2=\pi R^2.}$

This gives $2\pi \mathrm{Rr}=\pi R^2$ and so $2r=R$. The diameters of the 3 circles are $2r$, $4r$ and $6r$.