Air contains 20 % O2 by volume. How many cm³ of air will be required for oxidation of 100 cm³ of acetylene? Air contains 20 % $\ce{O2}$ by volume. How many cm³ of air will be required
for oxidation of 100 cm³ of acetylene?
(a) 1064 cm³
(b) 212.8 cm³
(c) 500 cm³
(d) 1250 cm³

Answer: (d)
My solution:
Equation: $\ce{2C2H2  +5O2 -&gt; 4CO2 +2H2O}$
From stoichiometric calculations:

2 moles of $\ce{C2H2}$ will react with 5 moles of oxygen
1 $\ce{C2H5} \rightarrow $ 5/2 moles of $\ce{O2}$
$\pu{22.4L}$ $\ce{C2H2} \rightarrow 5/2\times \pu{22.4L}$ of $\ce{O2}$
1 volume of air $\rightarrow$ 0.2 (20/100) volume of oxygen

Now the only thing I don’t know is how to convert the volume into liters so that I can find the correct answer.
Let's begin with your answer:
$2\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5\ \mathrm{mol}$ of oxygen.
$1\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5/2\ \mathrm{mol}$ of $\ce{O2}$.
If we suppose that the gases are under standard conditions, then
$22.4\ \mathrm{L}$ $\ce{C2H2}$ will react with  $\frac{5}{2} \times 22.4\ \mathrm{L  } ~ \ce{O2}$.
This means that:
$0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\ \mathrm{L}$ $\ce{O2}$.
$0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\times 5\ \mathrm{L}$ of air.
So, the volume of air is $1.25\ \mathrm{L}= 1250\ \mathrm{cm^3}$