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The combustion reaction of methane is:
$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

Here, the combustible gas is methane ($\text{CH}_4$).
Number of moles of $\text{CH}_4, n = 100\text{ mol}$

At standard temperature and pressure ($0^\circ\text{C}$ and $760\text{ mm Hg}$):
Volume of $1\text{ mole}$ of gas = $22.4\text{ L}$ (or $22.414\text{ L}$)

Therefore, the volume of the combustible gas ($\text{CH}_4$) is:
$V = 100\text{ mol} \times 22.414\text{ L/mol} = 2241.4\text{ L} = 2.2414\text{ m}^3$

(Note: If the required volume of air for combustion was intended:
Volume of $\text{O}_2$ required $= 2 \times 2241.4\text{ L} = 4482.8\text{ L}$.
Since air contains $21\%\text{ v/v } \text{O}_2$, the volume of air required $= \frac{4482.8}{0.21} \approx 21346.7\text{ L} = 21.35\text{ m}^3$)