1. Use the conductivity cell relation

For a conductivity cell,

$\kappa = \frac{\text{cell constant}}{R}$

So,

$\text{cell constant} = \kappa R$

2. Find the cell constant using KCl solution

Given for KCl:

• Conductivity, $\kappa_{\text{KCl}} = 0.14\ \text{S m}^{-1}$
• Resistance, $R_{\text{KCl}} = 4.19\ \Omega$

Hence,

$\text{cell constant} = 0.14 \times 4.19 = 0.5866\ \text{m}^{-1}$

3. Use the same cell for HCl solution

For HCl:

• Resistance, $R_{\text{HCl}} = 1.03\ \Omega$

Therefore,

$\kappa_{\text{HCl}} = \frac{\text{cell constant}}{R_{\text{HCl}}} = \frac{0.5866}{1.03}$

$\kappa_{\text{HCl}} \approx 0.5695\ \text{S m}^{-1}$

4. Express in the required form

We need

$\kappa_{\text{HCl}} = \underline{\hspace{1cm}} \times 10^{-2}\ \text{S m}^{-1}$

Now,

$0.5695 = 56.95 \times 10^{-2}$

Rounded to the nearest integer:

$57 \times 10^{-2}\ \text{S m}^{-1}$

5. Final answer

The required integer is:

$\boxed{57}$