= Solution
The disc carries <mass> $M_d=2M/3$ and the thin rim carries $M_r=M/3$. About the wheel's central normal axis, add the <moment of inertia of a uniform disc> and the thin rim's <moment of inertia>:
$$
\boxed{I_z=\frac12M_dR^2+M_rR^2=\frac12\frac{2M}{3}R^2+\frac M3R^2=\frac23MR^2.}
$$
Treat the wheel as a thin <planar lamina>, as implied by the disc-and-thin-rim model. Interpret the requested perpendicular axis as perpendicular to the wheel's original axis and tangent to the rim: if the wheel lies in the $xy$ <plane>, choose the tangent line $x=R,z=0$, parallel to the $y$ diameter. By <rotational symmetry>, the two diametral <moments of inertia> are equal. The <perpendicular axis theorem> therefore gives
$$
I_x=I_y=\frac{I_z}{2}=\frac13MR^2.
$$
The <parallel axis theorem> then gives the <tangent-axis inertia of an axisymmetric lamina>:
$$
\boxed{I_{\mathrm{tangent,in\ plane}}=I_y+MR^2=\frac43MR^2.}
$$
The axis direction matters. If “perpendicular” were intended to mean perpendicular to the wheel's plane, an axis through a rim point would instead be parallel to the original wheel axis, giving $I_{\mathrm{rim,normal}}=I_z+MR^2=5MR^2/3$. The $4MR^2/3$ answer uses a tangent axis perpendicular to the original axis; the $5MR^2/3$ value belongs to the different, explicitly identified normal axis. Finite wheel thickness would require additional information for the in-plane <moment of inertia>.
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