I will give a sketch of the solution because the detailed solution involves complicated algebra. It is just "brute force", and straightforward

a)moment of inertia about axis of rotation =(1/2)(MR²+mr²+2mR²)=I

at equilibrium position, kinetic energy =decrease in potential energy=mgR(1-cosθ)=(1/2)Iω²=(1/2)Iv²/R², upon doing the simplification you will arrive at the answer given

b)Torque =T =I*alpha=mgRθ=-I(ω^2)θ,ω²= mgR/I, T=2π/ω