Interpret the arrangement

Four identical discs of mass $M$ and diameter $a$ are placed in a plane in a $2 \times 2$ arrangement, touching each other.

Hence, each disc has radius $r = \frac{a}{2}.$

The axis $OO'$ lies in the plane of the discs and passes vertically through the middle of the arrangement, as is standard for this figure.

Distance of each disc’s center from the axis $OO'$

Since two discs fit side by side and touch each other, the horizontal distance between centers of neighboring discs is equal to the diameter $a$.

Therefore, the vertical axis through the middle is at a distance $d = \frac{a}{2}$ from the center of each disc.

Moment of inertia of one disc about a diameter in its own plane

For a uniform disc, the moment of inertia about a diameter is $I_{\text{diameter}} = \frac{1}{4}Mr^2.$

Since $r = \frac{a}{2}$, $I_{\text{cm, in-plane}} = \frac{1}{4}M\left(\frac{a}{2}\right)^2 = \frac{Ma^2}{16}.$

Apply parallel axis theorem for one disc

The required axis $OO'$ is parallel to this diameter axis and displaced by distance $d = \frac{a}{2}$.

So for one disc, $I_1 = I_{\text{cm}} + Md^2$ $I_1 = \frac{Ma^2}{16} + M\left(\frac{a}{2}\right)^2$ $I_1 = \frac{Ma^2}{16} + \frac{Ma^2}{4}$ $I_1 = \frac{5}{16}Ma^2.$

Moment of inertia of all four discs

Since all four contribute equally, $I = 4 \times \frac{5}{16}Ma^2 = \frac{5}{4}Ma^2.$

Compare with given form

Given, $I = \frac{x}{4}Ma^2.$

Therefore, $\frac{x}{4}Ma^2 = \frac{5}{4}Ma^2$ which gives $x = 5.$

Comparison with stored answer

The derived answer is $5$, whereas the stored correct answer is $3$.

The value $3$ would arise if one incorrectly used the moment of inertia of a disc about an axis perpendicular to its plane through the center, instead of the in-plane diameter axis required here. Hence the stored answer appears inconsistent with the standard interpretation of the figure and axis $OO'$ in the plane.