In the above reaction the Ethyl group of grignard reagent acts as a strong base.

As per the stoichiometric equation the number of moles of methanol is equal to the number of moles of ethane.

Volume of methanol$=$Volume of ethane$=$$2.24\mathrm{ml}$

Hence, number of moles of ethane

$\mathrm{n}=\frac{2.24\times {10}^{-3}}{22.4}={10}^{-4}$

Weight of the gas= Number of moles$\times$ Molecular mass

$\mathrm{W}=\mathrm{n}\times \mathrm{M}$

$={10}^{-4}\times 30=3\mathrm{mg}$