Four identical discs each of mass ' M ' and diameter ' a ' are...

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

Moment of inertia of disc about diagonal is ${I}_{1} = {I}_{3} = \dfrac{M{R}^{2}}{4}$

Now using parallel axis theorem,

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

${I}_{2} = {I}_{4} = \dfrac{M{R}^{2}}{4} + M{R}^{2} = \dfrac{5}{4}M{R}^{2}$

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

$I = {I}_{1} + {I}_{2} + {I}_{3} + {I}_{4}$

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

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

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

$I = \dfrac{3M{a}^{2}}{4}$, so,$x = 3$