= Solution
Use the usual dumbbell geometry: the uniform thin rod joins the disc centres and is perpendicular to both disc faces. The required axis through the rod midpoint lies in a direction parallel to a diameter of each disc. For either disc, <rotational symmetry> and the <perpendicular axis theorem> give
$$
I_{\rm diameter}=\frac12I_{\rm normal}=\frac14Ma^2.
$$
Each disc centre lies a distance $l/2$ from the requested axis. The <parallel axis theorem> therefore gives a disc contribution $Ma^2/4+M(l/2)^2$. The rod contributes
$$
I_{\rm rod}=\frac{m_0}{l}\int_{-l/2}^{l/2}x^2\,dx
=\frac1{12}m_0l^2.
$$
Add the two disc contributions and the rod's <moment of inertia>:
$$
\boxed{I=\frac1{12}m_0l^2+\frac12Ml^2+\frac12Ma^2.}
$$
Using $Ma^2/2$ for each disc's central contribution would instead use an axis perpendicular to its face, which is parallel to the rod and is not the requested axis.
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