OA Explanation:
(Refer to the figure below)

Let the radius of new cirlces (formed by cutting the plance H units above the centre O) be \(r\)

Now, combined surface area of these two pieces is 25% more than that of the sphere i.e. \(1.25(4πR^2)\)
Extra Area = \(0.25(4πR^2) = πR^2\)

This means, area of 2 new cirlces (centre at D and shaded) with radius \(r\) have combined area of \(πR^2\)
Also, area of new circle with radius \(r\) must be equal to \(\frac{ πR^2}{2}\)
i.e. \(πr^2 = \frac{ πR^2}{2}\)
or \(R = r\sqrt{2}\)

In △ODC;
\(R^2 = r^2 + H^2\)
\((r\sqrt{2})^2 = r^2 + H^2\)
\(H^2 = r^2\)
\(H = r\)

Col. A: \(r\)
Col. B: \(\frac{R}{\sqrt{6}} = \frac{(r\sqrt{2})}{\sqrt{6}}\)

Col. A: \(r\)
Col. B: \(\frac{r}{\sqrt{3}}\)

Multiplying both sides by \(\sqrt{3}\);

Col. A: \(r\sqrt{3}\)
Col. B: \(r\)

Col. A > Col. B
Hence, option A