Given balanced equation is

$\stackrel{⊝}{60\mathrm{H}}+\stackrel{⊝}{\mathrm{C}}\mathrm{l}⟶{\mathrm{ClO}}_3^{-}+3{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}$

$\to 10 {\mathrm{gKClO}}_3\Rightarrow \frac{10}{122.6} {\mathrm{molKCO}}_3$ in obtained

$\to$ from the above reaction, it is concluded that by
$6 \mathrm{F}$ charge $1 {\mathrm{molKClO}}_3$ is obtained.

$\to$ By the passage of $6 \mathrm{F}$ charge $=1 {\mathrm{molKClO}}_3$

$\therefore$ By the passage of $\frac{\mathrm{x}\times 10\times 60\times 60}{96500}\mathrm{F}$ charge

$=\frac{1}{6}\times \frac{\mathrm{x}\times 10\times 60\times 60}{96500}$

Now $\frac{\mathrm{x}\times 10\times 60\times 60}{6\times 96500}=\frac{10}{122.6}$

$\Rightarrow \mathrm{x}=\frac{10\times 965}{60\times 122.6}=\frac{965}{735.6}=1.311\simeq 1$

OR

$\mathrm{W}=\frac{\mathrm{E}}{\mathrm{F}}\times \mathrm{I}\times \mathrm{t}$

$10=\frac{122.6}{96500\times 6}\times \mathrm{x}\times 10\times 3600$

$\mathrm{X}=1.311$