In transforming 0.01 mole of $\mathrm{PbS}$ to ${\mathrm{PbSO}}_4$ the volume of 10 volume ${\mathrm{H}}_2{\mathrm{O}}_2$ required is 

$\mathrm{PbS}+4{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{PbSO}}_4+4{\mathrm{H}}_2\mathrm{O}$

$4{\mathrm{H}}_2{\mathrm{O}}_2\to 4{\mathrm{H}}_2\mathrm{O}+2{\mathrm{O}}_2$

 4 x 34                    2 mole

                              2 x 22.4 litre at STP = 44.8

For 1 mole of PbS     -----------   44.8  lit of ${\mathrm{O}}_2$ at STP is required

  0.01 mole of PbS      -----------   0.448 lit of $O_2$ at STP is required

              Volume of ${\mathrm{O}}_2$ at STP = Volume in ml x volume strength

                           448 ml          = Volume in ml x 10

                        Volume in ml  = $\frac{448}{10}=44.8$  ml