The density of $\ce{H2}$ is such that $1$ mole, or $2$ g $\ce{H2}$ occupies a volume $22.4$ L at $0$°C at $1$ atm. Its density is $\pu{\frac{2 g}{22.4 L} = 0.0892 g/L}$. Doing the same calculation with $\ce{CH4}$ and $\ce{C2H6}$ gives densities equal to $\ce{0.6670}$ g/L for $\ce{CH4}$ and $\pu{1.34}$ g/L for $\ce{C2H6}$

The density of the gas mixture is $14.4 · 0.0892 = 1.2857$ g/L. This value is between the densities of $\ce{CH4}$ and $\ce{C2H6}$. It means that these two gases can be mixed in such a proportion to give the observed density. So the choice a) can be an acceptable answer.

And indeed a mixture $10$% methane + $90$% ethane has the observed density.

The density of $\ce{H2}$ is such that $1$ mole, or $2$ g $\ce{H2}$ occupies a volume $22.4$ L at $0$°C at $1$ atm. Its density is $\pu{\frac{2 g}{22.4 L} = 0.0892 g/L}$. Doing the same calculation with $\ce{C2H2}$ and $\ce{C3H4}$ gives densities equal to $\ce{1.1607}$ g/L for $\ce{C2H2}$ and $\pu{1.7857}$ g/L for $\ce{C3H4}$

The density of the gas mixture is $14.4 · 0.0892 = 1.2857$ g/L. This value is between the densities of $\ce{C2H2}$ and $\ce{C3H4}$. It means that these two gases can be mixed to give the observed density. So the choice b) can be an acceptable answer.

And indeed a mixture $80$% $\ce{C2H2}$ + $20$% $\ce{C3H4}$ has the observed density.

The solution c) contains too heavy hydrocarbons. The solution d) contains a non-existant compound ($\ce{C4H9}$)