Given,
length of rod, $l=10 \,cm =0.1 \,m$
area of cross-section of rod, $A=2.8 \times 10^{-4}\, m ^{2}$
temperatuer at one end, $T_{1}=80^{\circ} C$
temperature at other end, $T_{2}=0^{\circ} C$
quantity of melted ice, $m =20 \,gm$
time taken to melt ice, $=5 \,min =300 \,sec$
and latent heat of ice, $s=80\, cal \,g ^{-1}$
Now, rate of the heat flow $=\frac{m \times s \times 4.184}{t}$
$=\frac{20 \times 80 \times 4.184}{300}=22.314 J / s$
Rate of heat flow in the rod, $\Delta Q=\frac{k A \cdot \Delta T}{l}$
$22.314=\frac{k\left(2.8 \times 10^{4}\right) \times 80}{0.1} $
$\therefore k=99.61 \approx 100\, Js ^{-1} \,m ^{-1} \,K ^{-1}$