I did not solve this problem, but I noticed one super important thing which might be valuable in this problem.

Let O be the center of first first with radius 2 cm.

Let C be the center of the second circle.

Draw OC, which we are told is 20 cm.

Draw External Tangent at A on circle O and B on circle C.

Draw AC which we know is 16 cm.

Let P be the intesection of OC and AB.

Then triangle OAP is similar to triangle PCB. *

*)Because <OAP = <PCB = 90 because it is tangent and hence perpendicular with radius. And <AP0 = <CPB because they are vertical angles.