Explanation

(b)
Let the oxidation number of oxygen in following compounds is $x.$
In ${OF}_{2}$
$x + ( - 1)2 = 0$
$x = \ + 2$
In ${KO}_{2}$
$+ 1 + (x \times 2) = 0$
$2x = - 1$
$x = - \frac{1}{2}$
In ${BaO}_{2}$
$+ 2 + (x \times 2) = 0$
$2x = - 2$
$x = - 1$
In $O_{3},\ oxidation\ number\ of\ oxygen\ is\ zero\ because\ oxidation\ number\ of\ an\ element\ in$
free state or in any of its allotropic form is always zero.
Thus, the increasing order of oxidation number is
${\ \ BaO}_{2} < {KO}_{2} < O_{3} < {OF}_{2}$
$- 1$ $- \frac{1}{2\ }$ 0 +2