#### Answer

$k=0.87m$

#### Work Step by Step

We can determine the required radius of gyration as follows:
$I=\Sigma (I^{\prime}+md^2)$
$\implies I=\frac{4(0.2)^2}{2}+4(1.2)^2+\frac{2(0.1)^2}{2}+2(0.6)^2+\frac{4(1.5)^2}{12}+4(0.5)^2=7.57Kgm^2$
and $m_p=4+2+4=10Kg$
Now $k=\sqrt{\frac{I}{m_p}}$
We plug in the known values to obtain:
$k=\sqrt{\frac{7.57}{10}}$
$\implies k=0.87m$