Interpret the figure

Seven identical disks are welded symmetrically in the usual hexagonal arrangement:

• one disk at the center,
• six surrounding it, each touching the central disk.

The point $P$ is at the outermost edge of one of the outer disks, and the axis is perpendicular to the plane through $P$.

Moment of inertia of one disk about its own center

For a uniform circular disk, about an axis normal to its plane through its center: $I_{\text{cm}} = \frac{1}{2}MR^2$

Use parallel axis theorem

For each disk, $I_P = I_{\text{cm}} + Md^2 = \frac{1}{2}MR^2 + Md^2$ where $d$ is the distance from the center of that disk to point $P$.

Locate centers and compute distances from $P$

Let the center of the middle disk be $O=(0,0)$ and radius be $R$.

The six outer centers are at distance $2R$ from $O$, at angles $0^\circ,60^\circ,120^\circ,180^\circ,240^\circ,300^\circ$.

Choose the rightmost outer disk center at $C_1=(2R,0)$ Then point $P$, being the outermost point on this disk, is at $P=(3R,0)$

Now compute $d$ for all seven disks:

• Disk 1: right outer disk $d_1 = R$

• Disk 2: center disk $d_2 = 3R$

• Disk 3: left outer disk at $(-2R,0)$ $d_3 = 5R$

• Disk 4 and 5: at $(R,\sqrt{3}R)$ and $(R,-\sqrt{3}R)$ $d^2 = (3R-R)^2 + (\sqrt{3}R)^2 = (2R)^2 + 3R^2 = 7R^2$ so each has $d=\sqrt{7}R$

• Disk 6 and 7: at $(-R,\sqrt{3}R)$ and $(-R,-\sqrt{3}R)$ $d^2 = (3R+R)^2 + (\sqrt{3}R)^2 = (4R)^2 + 3R^2 = 19R^2$ so each has $d=\sqrt{19}R$

Add moments of inertia of all disks

Total moment of inertia: $I = \sum \left(\frac{1}{2}MR^2 + Md^2\right)$

Since there are 7 disks, $I = 7\cdot \frac{1}{2}MR^2 + M\sum d^2$

Now, $\sum d^2 = R^2 + 9R^2 + 25R^2 + 2(7R^2) + 2(19R^2)$ $= (1+9+25+14+38)R^2 = 87R^2$

Therefore, $I = \frac{7}{2}MR^2 + 87MR^2$ $= \left(\frac{7}{2} + 87\right)MR^2$ $= \frac{181}{2}MR^2$

Match with the options

$\boxed{I = \frac{181}{2}MR^2}$

So the correct option is A.