1. Interpret the figure and axis

   We have three identical discs, each of mass $M$ and radius $R$:

   • central disc $D_1$
   • two side discs $D_2$ and $D_3$, attached rigidly at opposite ends of $D_1$

   The axis $OO'$ passes through the centre of $D_1$ in the plane of the discs, as shown.

   Since the side discs touch the central disc externally, the distance of each side disc’s centre from the centre of $D_1$ is $d = R+R = 2R.$

2. Moment of inertia of the central disc $D_1$ about $OO'$

   For a disc, the moment of inertia about any diameter (axis in its plane through centre) is $I_{ ext{diameter}} = \frac{1}{4}MR^2.$

   Therefore, $I_1 = \frac{1}{4}MR^2.$

3. Moment of inertia of each side disc about $OO'$

   For each of $D_2$ and $D_3$, the axis $OO'$ is parallel to a diameter through its own centre, but shifted by distance $2R$.

   So using the parallel axis theorem: $I = I_{\text{through centre, parallel axis}} + Md^2.$

   Here,

   \qquad d=2R.$$ Hence for each side disc, $$I_2 = I_3 = \frac{1}{4}MR^2 + M(2R)^2 = \frac{1}{4}MR^2 + 4MR^2 = \frac{17}{4}MR^2.$$

4. Total moment of inertia

   $I_{\text{total}} = I_1 + I_2 + I_3$ $= \frac{1}{4}MR^2 + \frac{17}{4}MR^2 + \frac{17}{4}MR^2$ $= \frac{35}{4}MR^2.$

5. Compare with the given options

   The computed value is $\boxed{\frac{35}{4}MR^2}.$

   This does not match any of the listed options $A,B,C,D$.

6. Check possible interpretation issue

   If instead the intended axis were the one perpendicular to the plane of the discs through the centre of $D_1$, then:

   • for $D_1$: $I_1 = \frac{1}{2}MR^2$
   • for each side disc: $I = \frac{1}{2}MR^2 + M(2R)^2 = \frac{1}{2}MR^2 + 4MR^2 = \frac{9}{2}MR^2$ so $I_{\text{total}} = \frac{1}{2}MR^2 + 2\left(\frac{9}{2}MR^2\right) = \frac{19}{2}MR^2,$ which also does not match the options.

Therefore, based on the standard geometry and axis description, the stored answer $A: 3MR^2$ appears incorrect or the figure/axis in the problem statement is incomplete or inconsistent.