At $25°\mathrm{C}$ , the water so-obtained is in the liquid state, which is supposed to occupy negligible volume or almost zero volume.

Total volume of ${\mathrm{H}}_2$ and ${\mathrm{O}}_2$ before reaction
$=100\mathrm{m}\mathrm{L}+500\mathrm{m}\mathrm{L}=600\mathrm{m}\mathrm{L}$
and volume of ${\mathrm{H}}_2$ and ${\mathrm{O}}_2$ reacted = 200 mL + 100 mL = 300 mL
Volume of ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ formed = 0.00 mL
So decrease in volume $=(600-300+0.00)\mathrm{m}\mathrm{L}=300\mathrm{m}\mathrm{L}$