Correct Answer: $\sqrt{11}$

Solution :
Given: The radius of the outer circle is 6 cm and the chord PQ of the length 10 cm is a tangent to the inner circle.
Use the theorems,
Any point on a circle has a tangent that is perpendicular to the radius through the point of contact.
The perpendicular line bisects the chord.
Let $C_1$ and $C_2$ be the two circles having the same centre O.
And, PQ is a chord that touches the $C_1$ at point R.
Since, $OR \perp PQ$. So, PR = RQ = $\frac{10}{2}$ = 5 cm [The perpendicular line OR bisects the chord PQ].
In right angled $\triangle PRO$, using Pythagoras's theorem,
$(PO)^2=(PR)^2+(OR)^2$
⇒ $(OR)^2=6^2–5^2$
⇒ $(OR)^2=36–25$
⇒ $(OR)^2=11$
⇒ $OR=\sqrt{11}$ cm
Hence, the correct answer is $\sqrt{11}$.