Given here, 

Moment of inertia of a disc of mass $M$ and radius '$R$' about any of its diameter is $\frac{M{R}^2}{4}$. 

Moment of inertia of disc about centre is $I_{cm}=\frac{M{R}^2}{4}+\frac{M{R}^2}{4}=\frac{M{R}^2}{2}$

Using parallel axis theorem, we can write

$I=I_{cm}+M{d}^2$

$=\frac{M{R}^2}{2}+M{R}^2$

$=\frac{3}{2}M{R}^2$

Hence, the value of $x=3$.