==> $\frac{{{K_S}A({\theta _1} - {\theta _2})}}{{2x}} = \frac{{2KA({\theta _1} - \theta )}}{x}$
$\because {K_S} = \frac{{2 \times 2K \times K}}{{(2K + K)}} = \frac{4}{3}K$ and $({\theta _1} - {\theta _2}) = 36^\circ $
==> $\frac{{\frac{4}{3}KA \times 36}}{{2x}} = \frac{{2KA({\theta _1} - \theta )}}{x}$
Hence temperature difference across wall $A$ is $({\theta _1} - \theta ) = {12^o}C$
