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(C) In steady state, the rate of heat flow $(H)$ through each rod is the same because they are connected in series.The rate of heat flow is given by $

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$5/3$

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(C) In steady state, the rate of heat flow $(H)$ through each rod is the same because they are connected in series.The rate of heat flow is given by $H = \frac{KA(T_{high} - T_{low})}{L}$.Since the rods are identical, $A$ and $L$ are the same for all.For the first rod: $H = \frac{(2K)A(3T - T_1)}{L}$For the second rod: $H = \frac{KA(T_1 - T_2)}{L}$For the third rod: $H = \frac{(2K)A(T_2 - T)}{L}$Equating the heat flow rates:$2(3T - T_1) = (T_1 - T_2) = 2(T_2 - T)$From $2(3T - T_1) = (T_1 - T_2)$:$6T - 2T_1 = T_1 - T_2 \implies 3T_1 - T_2 = 6T$  --- $(1)$From $(T_1 - T_2) = 2(T_2 - T)$:$T_1 - T_2 = 2T_2 - 2T \implies T_1 - 3T_2 = -2T$  --- $(2)$Multiplying $(2)$ by $3$: $3T_1 - 9T_2 = -6T$ --- $(3)$Adding $(1)$ and $(3)$: $(3T_1 - T_2) + (3T_1 - 9T_2) = 6T - 6T$$6T_1 - 10T_2 = 0 \implies 6T_1 = 10T_2$Therefore, $T_1/T_2 = 10/6 = 5/3$.

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