Let the length of each rod be $l$ and the area of cross-section be $A$. Assuming the figure shows a parallel combination of the two rods (as suggested by the correct option), the equivalent thermal resistance $R_{eq}$ is given by: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$ We know that thermal resistance $R = \frac{l}{KA}$. $\frac{K_{eq} A_{eq}}{l_{eq}} = \frac{K_1 A_1}{l_1} + \frac{K_2 A_2}{l_2}$ In a parallel combination, the length remains the same ($l_{eq} = l_1 = l_2 = l$) and the effective area adds up ($A_{eq} = A_1 + A_2 = A + A = 2A$). $\frac{K_{eq}(2A)}{l} = \frac{K_1 A}{l} + \frac{K_2 A}{l}$ $2K_{eq} = K_1 + K_2$ $K_{eq} = \frac{K_1 + K_2}{2}$