Answer
$k_{\mathrm{B}} \lt k_{\mathrm{A}} \lt k_{\mathrm{D}} \lt k_{\mathrm{C}}$
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 $k$ is inverse to the ranking of $\Delta T.$
$\Delta T_{C}=2^{o} < \Delta T_{D}=3^{o} < \Delta T_{A}=5^{o} < \Delta T_{\mathrm{B}}=10^{o}$
so,
$k_{\mathrm{B}} \lt k_{\mathrm{A}} \lt k_{\mathrm{D}} \lt k_{\mathrm{C}}$
