 Solution

The moment of inertia of a uniform thin ring about its perpendicular central axis is $M_a{a}^2$. Its two in-plane diameters are equivalent by rotational symmetry, so the perpendicular axis theorem gives $M_a{a}^2/2$ about either diameter. The central rod contribution about an axis perpendicular to its length is $M_l{l}^2/12$.

The requested axis is parallel to these central axes and passes through the whole racquet's centre of mass. Applying the parallel axis theorem separately givesAn equivalent reduced expression isBoth refer to the in-plane axis perpendicular to the handle; the ring contribution would be different for an axis normal to the racquet face.