Dear Student

Active Mass = $\frac{No. of moles}{Volume in\hspace{0.17em}L}$
So,
Molar Mass of CO₂ = 44 g/mol
Molar Mass of H₂ = 2 g/mol
Molar Mass of N₂ = 28 g/mol

​Given -
Mass of C₂ = 22 g
Mass of H₂ = 3 g
Mass of N₂ = 7 g

We know,
No. of moles = $\frac{Mass}{Molar Mass}$ 
So, 
Active Mass = $\frac{Mass}{Molar Mass x Volume in L}$

Let the volume of the gaseous mixture be 'V' litres
 
Active Mass of CO₂ = $\frac{22}{44 x V}$ = $\frac{1}{2V}$ = $\frac{0.5}{V}$

Active Mass of ​H₂ = $\frac{3}{2 x V}$ = ​$\frac{1.5}{V}$

Active Mass of N₂ = $\frac{7}{28 x V}$ = $\frac{0.25}{V}$

Ratio will be 0.5 : 1.5 : 0.25 or 1 : 3 : 0.5

So, the correct option is (4).

Regards