[tex]\mathrm{CO}_2[/tex] has 2 oxygens for every carbon. [tex]\mathrm{CO}[/tex] will contribute one of each for every molecule. You then have to have an even multiple of [tex]\mathrm{CO}[/tex] molecules to not end up with an odd number of oxygens.
So if you take [tex]2\mathrm{CO}[/tex], you should end up with 2 carbons on the RHS. [tex]2\mathrm{CO}_2[/tex] then has 4 oxygens. 2 of them are already accounted for, so you just need one [tex]\mathrm O_2[/tex] to balance out the reaction.
[tex]1\mathrm O_2+2\mathrm{CO}\Rightarrow2\mathrm{CO}_2[/tex]
