Let 1 mol of gas is present. So, 22.4 L will be the volume of the mixture

$$\begin{gathered} \mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{volume}} \\ 0.543=\frac{\mathrm{Mass}}{22.4}=12.16 \mathrm{g} \end{gathered}$$

Molar mass of mixture is 12.16 g.

% of oxygen present in mixture = x

% of He present in mixture = 1-x

Molar Mass of oxygen and He are 32 and 4 g/mol respectively

$$\begin{gathered} 12.16=32\times \mathrm{x}+(1-\mathrm{x})4 \\ 12.16=32\mathrm{x}+4-4\mathrm{x} \\ 8.16=\mathrm{x}\left(32-4\right) \\ 8.16=28\mathrm{x} \\ \mathrm{x}=0.29 \end{gathered}$$

From partial pressure of O₂, the mass of oxygen is determined and the percentage of oxygen becomes 80%.