Answer
$R = 0.841~R_0$
Work Step by Step
Let $R$ be the radius of the piece that is removed.
Let $M$ be the mass of this piece.
$M = \frac{\pi R^2}{\pi R_0^2}~M_0$
$M = \frac{R^2}{R_0^2}~M_0$
The moment of inertia of the piece that is removed should be half of the original moment of inertia. We can find the radius $R$.
$\frac{1}{2}MR^2 = \frac{1}{2}\times \frac{1}{2}M_0R_0^2$
$(\frac{R^2}{R_0^2}~M_0)R^2 = \frac{1}{2}M_0R_0^2$
$R^4 = \frac{1}{2}R_0^4$
$R = (\frac{1}{2})^{1/4}R_0$
$R = 0.841~R_0$
