Moment of inertia of the hole about the large disk center if it were filled ...

$I_{hole}=\dfrac{1}{2}\cdot \dfrac{M}{16} \cdot \left(\dfrac{R}{4}\right)^2 + \dfrac{M}{16} \cdot \left(\dfrac{R}{2}\right)^2$

$I_{hole}= \dfrac{MR^2}{512} + \dfrac{8MR^2}{512} = \dfrac{9MR^2}{512}$

$I_{disk \, with \, hole} + I_{hole} = I_{disk}$

$I_{disk \, with \, hole} = I_{disk} - I_{hole}$

$I_{disk \, with \, hole} = \dfrac{MR^2}{2} - \dfrac{9MR^2}{512}$

$I_{disk \, with \, hole} = \dfrac{256MR^2-9MR^2}{512} = \dfrac{247MR^2}{512}$

... the Chipster beat me to it.