Correct Answer: $21-14\sqrt{2}$

Solution :

We have $PQRS$ as a rectangle and a semicircle with $SR$ as diameter and $QR = 7\;\mathrm{cm}$.
Construction: Draw a line from $Q$ on line $ST$ (where $T$ is the centre of the semicircle) and pass through point $U$ (where $U$ is the point of contact and $O$ is the centre of the circle).
Now, $RT$ and $QR$ are the radius of the bigger circle and equal to $7\;\mathrm{cm}$.
In $\triangle QRT$,
$QT^2=QR^2+TR^2$
$QT^2 = 49 + 49$
$ QT = 7\sqrt{2}$ ____(i)
Let the radius of the smaller circle $=r\;\mathrm{cm}$
$QO = r\sqrt{2}$
From the figure,
$QT = TU + UO + OQ$
$QT = 7 + r + r\sqrt{2}$ ____(ii)
From equation (i) and (ii)
$7\sqrt{2} = 7 + r (\sqrt{2} + 1)$
⇒ $r = \frac{7 (\sqrt{2} - 1)}{(\sqrt{2} + 1)}$
⇒ $r = 7 (3 - 2\sqrt{2})$
Hence, the correct answer is $21 - 14\sqrt{2}$.