b
(b) In series rate of flow of heat is same
==> $\frac{{{K_A}A({\theta _1} - \theta )}}{l} = \frac{{{K_B}A(\theta - {\theta _2})}}{l}$
==> $3{K_B}({\theta _1} - \theta ) = {K_B}(\theta - {\theta _2})$
==> $3({\theta _1} - \theta ) = (\theta - {\theta _2})$
==> $3{\theta _1} - 3\theta = \theta - {\theta _2}$==> $4{\theta _1} - 4\theta = {\theta _1} - {\theta _2}$
==> $4({\theta _1} - \theta ) = ({\theta _1} - {\theta _2})$
==> $4({\theta _1} - \theta ) = 20$==> $({\theta _1} - \theta ) = 5^\circ C$
