$6 {OH}^{-}+{Cl} ^{-}\rightarrow {ClO}_{3}^{-}+3 {H}_{2} {O}+6 {e}^{-}$

$\rightarrow 10\, {~g}\, {KClO}_{3} \Rightarrow \frac{10}{122.6} \,{~mol} \,{KCO} 3$ in obtained

$\rightarrow$ From the above reaction, it is concluded that by $6\, {~F}$ charge $1\, {~mol} \,{KClO}_{3}$ is obtained.

$\rightarrow$ By the passage of $6\, {~F}$ charge $=1\, {~mol} \,{KClO}_{3}$

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

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

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

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

OR

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

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

$X=1.311$