Four identical discs each of mass ' M ' and diameter ' a ' are arranged in a small plane as…

Considering four identical discs as $1, 2, 3$ and $4$.

Moment of inertia of disc about diagonal is $I_1=I_3=\frac{M{R}^2}{4}$

Now using parallel axis theorem, 

Moment of inertia of disc $2$ and $4$ is 

$I_2=I_4=\frac{M{R}^2}{4}+M{R}^2=\frac{5}{4}M{R}^2$

So, net moment of inertia about $O{O}^{\text{'}}$ is 

 $I=I_1+I_2+I_3+I_4$

$=2\left(\frac{M{R}^2}{4}\right)+2\left(\frac{5}{4}M{R}^2\right)$

 $=\frac{M{R}^2}{2}+\frac{5}{2}M{R}^2$$=3M{R}^2$

Putting $R=\frac{a}{2}$

$I=\frac{3M{a}^2}{4}$, so,$x=3$