1. Formula for radius of gyration

For a thin rod of length $L$, about an axis perpendicular to its length and passing through its center, the moment of inertia is

$I = \frac{1}{12}ML^2$

Radius of gyration $k$ is defined by

$I = Mk^2$

So,

$Mk^2 = \frac{1}{12}ML^2$

$k^2 = \frac{L^2}{12}$

$k = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$

2. Substitute the given length

Given,

$L = 10\sqrt{3}\,\text{m}$

Therefore,

$k = \frac{10\sqrt{3}}{2\sqrt{3}} = \frac{10}{2} = 5\,\text{m}$

3. Final answer

$\boxed{5}$

4. Comparison with stored answer

Stored correct answer = $5$

My derived answer is also $5$, so they agree.