Explanation:

(Refer to the figure below)

Since, AM = ME = \(R\)
AME is an isosceles right triangle i.e. AE = \(\sqrt{2}R\)

Also, CD = DE = \(r\)
CDE is an isosceles right triangle i.e. CE = \(\sqrt{2}r\)

AE = \(R + r + \sqrt{2}r\)
\(\sqrt{2}R = R + r + \sqrt{2}r\)
\((\sqrt{2}-1)R = (\sqrt{2}+1)r\)

Col. A: \(R\)
Col. B: \(4r\)

Dividing both sides with \(r\);

Col. A: \(\frac{R}{r}\)
Col. B: 4

Col. A: \(\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}\)
Col. B: 4

Col. A: 5.83
Col. B: 4

Hence, option A