Eudiometry

10 ml hydrocarbon + 33 ml O2 -> 28 ml of CO2 and H2O. CO2 + KOH -> K2CO3. All the CO2 is consumed. Then, we have 8 ml of H2O. 28 ml (CO2 + H2O) - 8 ml H2O = 20 ml CO2

In its essence, this is a trick question. There is a lot of useless data.

After cooling, the residual gas had a volume of $28 \, mL$ . On treatment with $KOH$ , the volume decreased to $8 \, mL$ .

Potassium Hydroxide ( $KOH$ ) absorbs $CO_2$ . Hence, the volume of $CO_2$ is $(28-8) \, mL = \underline{20 \, mL}$

KOH absorbed CO2 and 20 mL residue directly implies answer is 20 mL

I was kinda surprised when I found out this question was so easy. The data given is more or less useless except for the volume before and after treatment with KOH. CO2= 28-8=20ml.....