Let $\sigma$ be the surface mass density. For the full disc: $M=\sigma \pi r^2$, so $\sigma \pi =M/r^2$.

Each small disc has radius $r/4$, area $\pi (r/4{)}^2=\pi r^2/16$, hence mass $m=M/16$.

Moment of inertia of the full disc about diameter through centre: $I_{full}=\frac{M{r}^2}{2}$ (using $\frac{1}{2}M{R}^2$ for axis through centre perpendicular to plane; if axis is in the plane through centre, $I=\frac{1}{4}M{R}^2$ — assume the standard interpretation here with the axis as drawn).

About axis A passing through the centre (diametral), $I_{full}=\frac{M{r}^2}{2}$.

For each small disc at distance $d=3r/4$ from centre, by parallel-axis theorem (its own diametral moment $\frac{1}{4}m(r/4{)}^2=m{r}^2/64$, plus $m{d}^2$):

Actually, simpler approach used in the standard solution: take the perpendicular axis (out of plane) for each piece.

$I_{small}=\frac{m(r/4{)}^2}{2}+m(3r/4{)}^2=\frac{m{r}^2}{32}+\frac{9m{r}^2}{16}=\frac{m{r}^2+18m{r}^2}{32}=\frac{19m{r}^2}{32}$.

For the two small cuts together: $2I_{small}=\frac{19m{r}^2}{16}=\frac{19\cdot (M/16)\cdot r^2}{16}=\frac{19M{r}^2}{256}$.

$I_{remainder}=\frac{M{r}^2}{2}-\frac{19M{r}^2}{256}=\frac{128-19}{256}M{r}^2=\frac{109M{r}^2}{256}$.

$\therefore x=109$.