As per the given figure, two plates A and B of thermal conductivity...

The thermal resistance of rod 1 will be ${R}_{1} = \dfrac{{l}_{1}}{KA} = \dfrac{4}{KA}$ and rod 2 will be ${R}_{2} = \dfrac{{l}_{2}}{2KA} = \dfrac{2 . 5}{2KA}$.

Since they are connected in series, ${R}_{eq} = {R}_{1} + {R}_{2} = \dfrac{10 . 5}{2KA}$.

If equivalent thermal conductivity is ${K}_{eq}$, then ${R}_{eq} = \dfrac{{l}_{1} + {l}_{2}}{{K}_{eq}A} = \dfrac{6 . 5}{{K}_{eq}A}$. Therefore,

$\dfrac{10 . 5}{2KA} = \dfrac{6 . 5}{{K}_{eq}A} ⇒ {K}_{eq} = \dfrac{26}{21}K = \left(1 + \dfrac{5}{21}\right)K$

Hence, $\alpha = 21$.