Moment Of Inertia of rod about the fixed point is ML^2/3
Moment Of Inertia of bob about the fixed point is mL^2
now dispace the system by an angle A
so we know
Iα=torque
{(ML^2)/3 +mL^2}α = MgsinA*L/2 +mgsinA*L
as A is very small sinA=A
{(ML^2)/3 +mL^2}α =(Mg*L/2 +mg*L)A
so α is proportional to A so it is an SHM
which gives ω^2=[(M+3m)L^2]/[3g(M+2m)]
now T=2π/w