Given reaction is,

$2{{\mathrm{MnO}}_4}^{-}+10{\mathrm{I}}^{-}+16{\mathrm{H}}^{+}\to 2{\mathrm{Mn}}^{2+}+5{\mathrm{I}}_2+8{\mathrm{H}}_2\mathrm{O}$

Disappearance of the ${\mathrm{MnO}}_4^{-}$ ion,

$\begin{array}{r}-\frac{1}{2}\frac{d\left[{\mathrm{MnO}}_4^{-}\right]}{dt}=+\frac{1}{5}\frac{d\left[I_2\right]}{dt}\end{array}$

So, the rate of appearance of ${\mathrm{I}}_2$ is:

$\frac{d\left(I_2\right)}{dt}=\frac{5}{2}\times 4.56\times {10}^{-3}$

$\Rightarrow \frac{d\left(I_2\right)}{dt}=1.14\times {10}^{-2}{\mathrm{Ms}}^{-1}$