Correct Answer: 2 cm

Solution :
Let x be the radius of the circle.
In the right-angled $\triangle$ ABC,
AC ² = AB ² + BC ² (by Pythagoras Theorem)
⇒ AC ² = 6 ² + 8 ²
⇒ AC ² = 36 + 64
⇒ AC ² = 100
$\therefore$ AC = 10
Now in quadrilateral OPBR, $\angle$B = $\angle$P = $\angle$R = 90°
Hence, $\angle$ROP = 90° (sum of all angles of a quadrilateral is 360°) and also OP = OR (each radius).
Let OPBR be a square with each side x cm.
So, BP = RB = x cm
Therefore, CR = (8 − x) and PA = (6 − x)
Since the tangents from an external point to a circle are equal in length,
​​​​​​$\therefore$ AQ = AP = (6 − x) and CQ = CR = (8 − x)
Now, AC = AQ + CQ
⇒ 10 = 6 − x + 8 – x
⇒ 10 = 14 − 2x
$\therefore$ x = 2 cm
Hence, the correct answer is 2 cm.