Correct Answer: 15.82 cm

Solution :
We know that a perpendicular from the centre of a circle to a chord bisects the chord into two equal parts.
$AP = \frac{1}{2}\times AB = \frac{1}{2}\times28 = 14$ cm
$CQ = \frac{1}{2}\times CD = \frac{1}{2}\times22 = 11$ cm
Let $OP$ be $x$.
So, $OQ=x+4$
Using Pythagoras theorem on $\triangle$OPA, we get,
$OA^{2} = OP^{2} + AP^{2}$
$⇒OA^{2} = x^{2} +14^{2}$
$⇒OA^{2} = x^{2} +196$
$⇒OA = \sqrt {x^{2} +196}$
$OA = OC =\sqrt {x^{2} +196}$ [$\because$ Both are radii of the circle]
Again, using Pythagoras theorem on $\triangle$OQC, we get,
$OC^{2} = OQ^{2} + CQ^{2}$
$⇒(\sqrt {x^{2} +196})^{2} = (x+4)^{2} + 11^{2}$
$⇒x^{2} +196 = x^{2}+16+8x+  121$
$⇒8x = 59$
$\therefore x=\frac{59}{8}=7.38$
So, the radius $OA=OC=\sqrt{(7.38)^2+196}=\sqrt{54.46+196}=\sqrt{250.46}=15.82$ cm
Hence, the correct answer is 15.82 cm.