1. Radius of gyration definition

For a body of mass $M$, if its moment of inertia about a given axis is $I$, then the radius of gyration $k$ is defined by

$I = Mk^2$

So,

$k = \sqrt{\frac{I}{M}}$

2. Moment of inertia of a solid sphere

For a solid sphere about any diameter,

$I = \frac{2}{5}MR^2$

Here, the sphere rotates about axis $PQ$, which is a diameter-like axis through its center.

3. Substitute into the radius of gyration formula

$k^2 = \frac{I}{M} = \frac{1}{M}\left(\frac{2}{5}MR^2\right)=\frac{2}{5}R^2$

Given

$R = 5\text{ cm}$

So,

$k^2 = \frac{2}{5}(5)^2 = \frac{2}{5}\cdot 25 = 10$

Hence,

$k = \sqrt{10}\text{ cm}$

The question says the radius of gyration is $\sqrt{x}$ cm, so

$x = 10$

4. Comparison with stored answer

Stored correct answer = $110$

But the derived value is clearly

$x = 10$

So the stored answer does not match the physics formula for a solid sphere about a diameter.