20 gm of steam at 100°C mixed with 40gm of ice at 0°C. The mass of steam uncondensed is 

The quantity of heat required to rise the temperature of 40 gm of ice at 0°C to water at 100° C
$\begin{array}{l}\mathrm{Q}={\mathrm{mL}}_{\mathrm{ice}}+\mathrm{ms\Delta t}\\ =\left(40\mathrm{x}80\right)+40\times 1\times 100\\ =7200\mathrm{cal}\to \left(1\right)\end{array}$
If m is the mass of condensed steam, the heat supplied by condensed steam
$\mathrm{Q}={\mathrm{mL}}_{\mathrm{S}\text{ team }}=\mathrm{mX}540\to \left(2\right)$
From equation (1) and (2),  $\mathrm{mX}540=7200\Rightarrow \mathrm{m}=\frac{40}{3}\mathrm{gm}$
The mass of uncondensed steam $=20-\frac{40}{3}=\frac{20}{3}\mathrm{gm}$