The moment of inertia of the complete disc about axis A is
Idisc=21Mr2=256128Mr2.
Each removed circular region has radius r/4 and mass mi=M/(r2/16r2)=M/16.
For each removed disc at distance 3r/4 from axis A, using the parallel axis theorem:
Iremoved=21mi(r/4)2+mi(3r/4)2=512Mr2+2569Mr2=51219Mr2.
For two removed discs:
I2removed=51238Mr2=25619Mr2.
Therefore:
Iremainder=256128Mr2−25619Mr2=256109Mr2, so x=109.