The reaction:

${\mathrm{OCl}}^{\ominus }+{\mathrm{I}}^{- }\stackrel{{\mathrm{OH}}^{-}}{⟶}{\mathrm{OI}}^{\ominus }+{\mathrm{Cl}}^{\ominus }$

takes place in the following steps:

i. ${\mathrm{OCl}}^{\ominus }+{\mathrm{H}}_2\mathrm{O}\underset{k_2}{\stackrel{k_1}{\leftrightharpoons }}\mathrm{HOCl}+\mathrm{OH}$        (fast)

ii. ${\mathrm{I}}^{\ominus }+\mathrm{HOCl}\stackrel{k_3}{⟶}\mathrm{HOI}+{\mathrm{Cl}}^{\ominus }$            (slow)

iii. $\stackrel{\ominus }{\mathrm{O}}\mathrm{H}+\mathrm{HOI}\frac{{k_1}^{\mathrm{\prime }}}{{\mathrm{k}}_2^{\mathrm{\prime }}}{\mathrm{H}}_2\mathrm{O}+{\mathrm{OI}}^{\ominus }$        (fast)

The rate of consumption of I in the following equation is

 

Step (ii) is the rate-determining step.

$\therefore \text{ Rate }=k_3\left[{\mathrm{I}}^{\ominus }\right]\left[\mathrm{HOCl}\right]$

(HOCI) produced in step (i) is consumed in step (ii), so it is a reactive intermediate. Similarly, (HOI) produced in step (ii) is consumed in step (iii), so it is also a reactive intermediate. The concentration of both reactive species is determined from
their equilibrium constant value.

$\therefore$ from step (i)

$\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{\left[\mathrm{HOCl}\right]\left[\mathrm{OH}\right]}{\left[{\mathrm{OCl}}^{\ominus }\right]}\Rightarrow \left[\mathrm{HOCl}\right]=\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}\frac{\left[{\mathrm{OCl}}^{\ominus }\right]}{\left[\mathrm{OH}\right]}$   .. . (2) 

Substitute the value of [HOCI] from Eq. (2) in Eq. (1),

$\text{ Rate }=\frac{{\mathrm{k}}_1{\mathrm{k}}_3}{{\mathrm{k}}_2}\frac{\left[{\mathrm{OCl}}^{\ominus }\right]\left[{\mathrm{I}}^{\ominus }\right]}{\left[{\displaystyle \stackrel{⊝}{\mathrm{O}}}\mathrm{H}\right]}$

Hence answer is (3).