You'd need #"24.6 mL"# of oxygen for the combustion of #"CO"# and #"10.0 mL"# of oxygen for the combustion of hydrogen.
Let's start with the combustion of carbon monoxide. The balanced chemical equation is
#2CO_((g)) + O_(2(g))-> 2CO_(2(g))#
Since no mention of pressure or temperature is made, I'll assume we're at STP, where pressure is equal to 1 atm and temperature is equal to 273.15 K. Moreover, at STP, 1 mole of any ideal gas occupies 22.4 L, or #"22.4 dm"^3#. Since you have #"50 cm"^3# of carbon monoxide, the number of moles you have is
#n = V/V_("molar") = (50 * 10^(-3)L)/("22.4 L/mol") = 0.0022# #"moles"#
Notice that you have a #2:1# mole ratio between #"CO"# and #"O"_2#, which means that the number of oxygen moles you have is
#"0.0022 moles CO" * ("1 mole O"_2)/("2 moles CO") = 0.0011# #"moles"#
This translates into a volume of
#V = n_("oxygen") * V_("molar") = "0.0011 moles" * 22.4 "L"/"mole"#
#V = "0.0246 L" = 24.6# #"mL"#
This is the balanced chemical equation for the combustion of hydrogen
#2H_(2(g)) + O_(2(g)) -> 2H_2O_((g))#
The same strategy applies for this reaction as well. The number of hydrogen moles is
#n = V/V_("molar") = (20*10^(-3)"L")/("22.4 L/mol") = 0.00089# #"moles"#
Once again, we have a #1:1# mole ratio to apply, so the number of oxygen moles is
#"0.00089 moles H"_2 * ("1 mole O"_2)/("2 moles H"_2) = 0.00045# #"moles"#
Therefore, the volume of oxygen needed is
#V = n_("oxygen") * V_("molar") = "0.00045 moles" * 22.4 "L"/"mole"#
#V = "0.010 L" = 10.0# #"mL"#
