As external torque is zero, angular momentum remains conserved.
I1ω1+2I12ω1=(I1+2I1)ω
∴ Common angular velocity ω=65ω1
∴ Ef−Ei=21×(I1+2I1)ω2−21×I1ω12−21×2Il(2ω1)2
=21×23I1(65ω1)2−169I1ω12
=14475I1ω12−81I1ω12
=−1446I1ω12
=−24I1ω12