jmc

Using the diagram above, let the radius of $D$ be $3r$, and the radius of $E$ be $r$. Then, $EF=r$, and $CE=2-r$, so the Pythagorean theorem in $△CEF$ gives $CF=\sqrt{4-4r}$. Also, $CD=CA-AD=2-3r$, so$$DF=DC+CF=2-3r+\sqrt{4-4r}.$$Noting that $DE=4r$, we can now use the Pythagorean theorem in $△DEF$ to get$$(2-3r+\sqrt{4-4r}{)}^2+r^2=16r^2.$$ Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$. Notice that C, E and the point of tangency to circle C for circle E will be concurrent because C and E intersect the tangent line at a right angle, implying they must be on the same line.