When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., $0.001$ $M$. Then,

$Nal{O_3}\quad  \to \quad N{a^ + }\quad  + \quad lO_3^ – $

$0.001\,M$                                     $0.001\,M$

$Cu{\left( {Cl{O_3}} \right)_2}\quad  \to \quad C{u^{2 + }}\quad  + \quad 2ClO_3^ – $

$0.001\,M$                                                $0.001\,M$

Now, the solubility equilibrium for copper iodate can be Written as:

$Cu{\left( {l{O_3}} \right)_2}\quad  \to \quad Cu_{(aq)}^{2 + } + \quad 2l{O_{3\left( {aq} \right)}}$

Ionic product of copper iodate:

$=\left[ Cu ^{2+}\right]\left[10_{3}^{-}\right]^{2}$

$=(0.001)(0.001)^{2}$

$=1 \times 10^{-9}$

Since the ionic product $\left(1 \times 10^{-9}\right)$ is less than $K_{s p}\left(7.4 \times 10^{-8}\right),$ precipitation will not occur.