The composition of LPG is butane and isobutane. The amount of oxygen that would be required for combustion of $1 kg$ of LPG will be approximately

The combustion reaction of butane is as follows:

$\underset{1 mol}{C_4{H}_{10}( g)}+\frac{13}{2}{O}_2( g)\to 4{CO}_2( g)+5H_{2}O\left(l\right)$

From the above reaction we can say that (58g) of $C_4{H}_{10}$ will require $\left(\frac{13}{2}\times 32=208 g\right)$ of $O_2$

$\therefore O_2$ required for $1 kg$ butane

$=\frac{208}{58}\times 1000 g$

= $3.58 \times {10}^{3}g$

Hence the correct option is (D).