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Explanation:

AB = AK = 2

Quadrilateral ABCD is a square. 
∴ AC = 2√2

⇒ KC = AC - AK = 2√2 - 2

Let the centre of the smaller circle be O and let its radius be x. 

∴ OC = x√2 and KC = KO + OC = x + x√2

⇒ x + x√2 = 2√2 - 2

⇒ x = $\frac{2\sqrt{2}-2}{1+\sqrt{2}}$

⇒ x = $\frac{(2\sqrt{2}-2)}{(1+\sqrt{2})}\times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}$ = 6 - 4√2

Hence, option (d).