This problem was solved by Dhara from the University High School
  in Victoria, Australia. <br />
  <br />
  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 $$\pi (R+r)^2 -\pi R^2 -\pi r^2 = \pi R^2.$$
  <br />
  <br />
  This gives $2\pi Rr = \pi R^2$ and so $2r = R$. The diameters of
  the 3 circles are $2r$, $4r$ and $6r$. <br />