Answer
$m(KClO_{3})=4.4g$
Work Step by Step
$T=25^{\circ}C=(25+273.15)K=298.15K$
$p=762torr=762\times 133.3Pa = 101574.6Pa=101.5746kPa$
Let's suppose that $V$ is the volume of oxygen we want to generate. Since $25\%$ is waste, we are left with $0.75V$ of oxygen, which is equal to the total volume of $4$ 250ml-bottles.
$0.75V=4\times 250ml \implies 0.75V=1l \implies V=1.333l$
$pV(O_{2})=n(O_{2})RT \implies n(O_{2})=\frac{pV(O_{2})}{RT}=\frac{101.5746kPa\times 1.333dm^{3}}{8.314\frac{J}{K\times mol}\times 298.15K}=0.055mol$
According to the reaction of thermal decomposition of potassium chlorate:
$2KClO_{3}\rightarrow 2KCl+3O_{2}$,
$2mol$ of $KClO_{3}$ produce $3mol$ of oxygen. Therefore, we can obtain the following proportion:
$2mol:3mol=n(KClO_{3}):0.055mol$
$\implies n(KClO_{3})=\frac{2}{3}\times 0.055mol=0.036mol$
$\implies m(KClO_{3})=M(KClO_{3})\times n(KClO_{3})=122.5\frac{g}{mol}\times 0.036mol\approx4.4g$
