hi pulkit
i think this might help u
as we know that the moment of inertia of ring about the axis passing through its center and perpendicucal to the plane is MR2 and that of rod of length L about an aixs passing through its end and perpenducular to the length of rod is ML2/3
so by using these two bodies we can find the moment of inertia of this wheel ( considering rim as ring and spokes as rods )
we get
moment of inertia of rim as MR2 = 1*0.3*0.3 = 0.09 kg m2
and the moment of inertia of all spokes =50*ML2/3=50*0.01*0.3*0.3/3 =0.015 kg m2
so the total moment of inertia of the wheel about its axis is 0.09+0.015