1. Given electrode reaction

$6OH^- + Cl^- \to ClO_3^- + 3H_2O + 6e^-$

From the reaction, formation of 1 mole of $ClO_3^-$ requires transfer of 6 moles of electrons.

Since potassium chlorate is $KClO_3$, 1 mole of $KClO_3$ corresponds to 1 mole of $ClO_3^-$ formed.

2. Moles of $KClO_3$ produced

Given mass of $KClO_3 = 10.0\,g$

Molar mass of $KClO_3 = 122.6\,g\,mol^{-1}$

$n(KClO_3)=\frac{10.0}{122.6}=0.08157\,mol$

3. Moles of electrons required

From stoichiometry:

$1\text{ mol }KClO_3 \Rightarrow 6\text{ mol }e^-$

So,

$n(e^-)=6\times 0.08157=0.4894\,mol$

4. Total charge required

Using Faraday constant $F=96500\,C$:

$Q=n(e^-)F=0.4894\times 96500$

$Q\approx 47230\,C$

5. Use $Q = It$ to find current

Time $t=10\,h=10\times 3600=36000\,s$

$I=\frac{Q}{t}=\frac{47230}{36000}\approx 1.31\,A$

Nearest integer:

$x=1$

6. Comparison with stored answer

Derived answer = $1$

Stored correct answer = $1$

So, the derived answer agrees with the stored answer.