Let r and R be the radii of smaller and bigger circles respectively, then
OO' = R + r
⇒    R - r = 6 cm    ....(i)

It is also given that
Sum of the area of the two circles = 116 π cm²
⇒ πR² + πr² = 116 π
⇒ π(R² + r²) = 116 π
⇒    R² + r² = 116    ...(ii)
We know that
(R - r)² = R² + r² - 2Rr ...(iii)
Putting the values of (i) and (ii) in (iii), we get
(6)² = 116 - 2Rr
⇒    36 = 116 - 2Rr
⇒    2Rr = 80
⇒    Rr = 40
Now, (R + r)² = R² + r² + 2 Rr
= 116 + 2 × 40
[From (ii) R² + r² = 116]
= 116 + 80
= 196
⇒    R + r = 14    ....(iv)
Solving (i) and (iv), we get
R = 10 cm
and    r = 4 cm.