1. Write the electrode reaction and electron requirement

Given: $6OH^- + Cl^- \rightarrow ClO_3^- + 3H_2O + 6e^-$

This shows that formation of 1 mole of $ClO_3^-$ involves 6 moles of electrons.

So, for $1$ mole of $KClO_3$, electrons needed $= 6$ moles.

2. Calculate moles of $KClO_3$ to be produced

Mass required $= 10\,\text{g}$

Molar mass of $KClO_3 = 122\,\text{g mol}^{-1}$

$n(KClO_3)=\frac{10}{122}=0.08197\,\text{mol}$

3. Calculate moles of electrons required

$n(e^-)=6\times 0.08197=0.4918\,\text{mol}$

4. Convert electron moles to charge actually needed for the reaction

Using $1$ mole electrons $= F = 96500\,\text{C}$,

$Q_{\text{reaction}}=0.4918\times 96500=47459\,\text{C}$

5. Account for 60% current efficiency

Only $60\%$ of supplied current is utilized, so:

$Q_{\text{supplied}}=\frac{Q_{\text{reaction}}}{0.60}$

$Q_{\text{supplied}}=\frac{47459}{0.60}=79098\,\text{C}$

6. Use $Q = It$ to find time

Given current $I = 2\,\text{A}$,

$t=\frac{Q}{I}=\frac{79098}{2}=39549\,\text{s}$

Convert into hours:

$t=\frac{39549}{3600}=10.99\,\text{h}$

Rounded to nearest hour:

$\boxed{11}$

7. Comparison with stored answer

Derived answer = $11$

Stored correct answer = $11$

So they agree.