#### Answer

$k_x=57.7mm$

#### Work Step by Step

We can find the required radius of gyration as follows:
$k^2_x=\frac{I_{mx}}{m}$
We plug in the known values to obtain:
$k^2_x=\frac{\int_{0}^{200}\rho \pi \frac{250}{2}x^2 dx}{\int _0 ^{200}\rho \pi 50 x dx}$
After taking the integral and applying the limits, we obtain:
$k^2_x=\frac{\frac{50}{2}\frac{(200)^3}{3}}{\frac{(200)^2}{2}}$
This simplifies to:
$k^2_x=\frac{1000}{3}$
Which further simplifies to:
$k_x=57.7mm$