Correct Answer: 15 cm

Solution :
Given, AB = AC = 1$2\sqrt{5}$cm and BC = 24 cm
Join OB, OC, and OA.
Draw AD$\perp$ BC which will pass through centre O
OD bisects BC in D as perpendicular from the centre to a chord bisects the chord.
So, BD = CD = 12 cm
Using Pythagoras theorem in $\triangle$ ABD,
AB ² = AD ² + BD ²
Or, $(12\sqrt{5})^{2}$ = AD ² + 12 ²
Or, AD ² = 576
Or, AD = 24 cm
Let the radius of the circle be OA = OB = OC = r
So, OD = AD – AO = 24 – r
Using Pythagoras theorem in $\triangle$ OBD,
r ² =12 ² + (24 – r) ²
Or, r ² = 144 + 576 + r ² – 48r
Or, r = 15 cm
Hence, the correct answer is 15 cm.