Suppose there are n₁ moles of hydrogen and n₂ moles of helium in the given mixture. Then the pressure of the mixture will be 

​​​​$\text{P}=\frac{{\text{n}}_1\text{RT}}{\text{V}}+\frac{{\text{n}}_2\text{RT}}{\text{V}}=\left({\text{n}}_1+{\text{n}}_2\right)\frac{\text{RT}}{\text{V}}$

$2\times \text{101.3}\times {10}^3=\left({\text{n}}_1+{\text{n}}_2\right)\frac{\left(\text{8.3}\right)\left(300\right)}{20\times {10}^{-3}}$

$\left({\text{n}}_1+{\text{n}}_2\right)=\frac{2\times \text{101.3}\times 10^3\times 20\times 10^{-3}}{\left(\text{8.3}\right)\left(300\right)}=\text{1.62}$   ...(i)
The mass of the mixture is (in $\mathrm{g}$)

 ${\text{n}}_1\times 2+{\text{n}}_2\times 4=5$ 

$\left({\text{n}}_1+2{\text{n}}_2\right)=\text{2.5}$ ......(ii)

Solving Eqs. (i) and (ii), ${\mathrm{n}}_1 = 0.74$, ${\mathrm{n}}_2 = 0.88$

Hence  $\frac{{\text{m}}_{\text{H}}}{{\text{m}}_{\text{He}}}=\frac{\text{0.74}\times 2}{\text{0.88}\times 4}=\frac{\text{1.48}}{\text{3.52}}\to \text{2 : 5}$