Answer
$\Delta T_{\mathrm{B}} < \Delta T_{\mathrm{A}} < \Delta T_{\mathrm{C}}$
Work Step by Step
Since opposite sides of the composite slab are held at the fixed temperatures,
the heat flow, $\displaystyle \frac{Q}{t}$ through each slab is equal.
Expressing $\Delta T$ from 16-16, $Q=kA(\displaystyle \frac{\Delta T}{L})t$
$\displaystyle \Delta T=\frac{(\frac{Q}{t})\cdot L}{kA}$, and we are given that L and A are also constant.
Conclusion: $\Delta T$ is inversely proportional to $k,$ the thermal conductivity.
The ranking of $\Delta T$ is inverse to the ranking of $k.$
$\Delta T_{\mathrm{B}} < \Delta T_{\mathrm{A}} < \Delta T_{\mathrm{C}}$
