A cylinder of radius R made of material of thermal conductivity K1 …... Ans (K1+8K2)/9

Let the difference between the temperatures across the two ends be ΔT length of the cylinder be L.
Let ΔH1 and ΔH2 is the heat transferred by the inner and outer conductor in time Δt.
Heat transferred by inner cylinder in time Δt is given
ΔH1/ Δt = (K1πR2 ΔT)/L
=> ΔH1 = (K1πR2 ΔT)Δt/L ---1
Heat transferred by outer cylinder in time Δt is given
ΔH2/ Δt = (K2π{(2R)2 – R2 }ΔT)/L
=> ΔH2 = (3K2πR2 ΔT)Δt/L ---2
Total heat transferred
ΔH = ΔH1 + ΔH2
From eqn. 1 and 2
ΔH = (K1πR2 ΔT)Δt/L + (3K2πR2 ΔT)Δt/L
=> ΔH/ Δt = (K1 + 3K2)πR2/L
Now the total radius of the system R’ = 2R
ΔH/ Δt = (K1 + 3K2)π(2R)2/4L
= (K1 + 3K2)π(R’)2/4L
= 1/4(K1 + 3K2) π(R’)2/L
= K π(R’)2/L
K = equivalent thermal conductivity
K = 1/4(K1 + 3K2)