Correct Answer: $6+\sqrt{161}$

Solution :
Let PQ be a common chord. O and O' are the centres of the two circles such that
OP = 15 cm, O'P = 10 cm, and PQ = 16 cm
OO' perpendicularly bisects PQ at L.
So, PL = $\frac{1}{2}$ × PQ = $\frac{16}{2}$ = 8 cm
In $\triangle$ OLP, OP is hypotenuse
So, OP ² = OL ² + PL ²
⇒ OL = $\sqrt{OP^{2} - PL^{2}}$
= $\sqrt{15^{2} - 8^{2}}$
= $\sqrt{161}$ cm
Similarly, in $\triangle$ O'LP, O'P is hypotenuse
So, O'P ² = O'L ² + PL ²
⇒ O'L = $\sqrt{O'P^{2} - PL^{2}}$
= $\sqrt{10^{2} - 8^{2}}$
= $\sqrt{36}$
= 6 cm
So, OO' = O'L + OL
= 6 + $\sqrt{161}$
Hence, the correct answer is $6 + \sqrt{161}$.