Answer: Radius = \(4\sqrt{3}\) cm
OP ⊥ \(l\) (tangent ⊥ radius) and RS ∥ \(l\), so OQ ⊥ RS. The perpendicular from the centre bisects the chord, so RQ = 6.
Let the radius be r. Then OR = r and OQ = \(\frac{r}{2}\). Draw seg OR. In right-angled \(\triangle OQR\):
\[r^2 = \left(\frac{r}{2}\right)^2 + 6^2\] \[\frac{3r^2}{4} = 36\] \[r^2 = 48,\quad r = 4\sqrt{3}\]