The long slender rod is welded to the thin circular disc of radius $R=0.5 \mathrm{m}$ tangential in the plane of disc. Therefore, tangential axis of disc is along the rod.

The moment of inertia of disc along the tangential axis in the plane of disc is given by,

$I=\frac{5}{4}M{R}^2$

Radius of gyration is,

$K=\sqrt{\frac{I}{M}}=\sqrt{\frac{\frac{5}{4}M{R}^2}{M}}=\frac{\sqrt{5}}{2}R$

$\Rightarrow K=\frac{\sqrt{5}}{2}\times 0.5=\frac{\sqrt{5}}{4} \mathrm{m}$