$I_2+\mathrm{10 }HN{O}_3\to \mathrm{2 }HI{O}_3+\mathrm{10 }N{O}_2+\mathrm{4 }{H}_{2}O$

Iodine gets oxidized to $I{O}_3^{-}$ when it reacts with an oxidizing agent $\left(HN{O}_3\right).$ The oxidation number of $I$ in product will be

Oxidation: $I_2-I{O}_3^{-}$

$-1=x+3\times \left(-2\right)$

$-1=x-6$

$x=-1+6=+5$

Reduction: concentrated HNO₃ gives NO₂.