Combustion reaction

C \quad + \quad O_2 \rightarrow CO_2 \\12  kg \quad 32  kg

Weight of oxygen required to burn 12 kg C = 32 kg

Weight of oxygen required to burn 3 kg     C =\frac{32}{12} × 3 = 8 ~kg

As, air contains 23% oxygen by weight

Therefore, weight of air required = 8 × \frac{100}{23}= 34.783 ~kg = 34783 ~g

Volume of air required

as, 1 mole of any gas at STP occupies 22.4 L

Therefore, volume occupied by 1 mole air = 28.94 g air = 22.4 L

(molecular weight of air = 28.94 g)

\begin{aligned}\text{Volume occupied by 34783 g of air}=  & \frac{22.4}{28.94}  × 34783\\ \\=  & 26.92 × 10^3 ~L = 26.92  m^3\\  &  (\text{Since}  1 ~L = 10^3 ~m^3)\end{aligned}