1. Given data

• Concentration: $0.2\%\,(w/v)$ NaOH
• Resistivity: $870.0\,\text{m}\Omega\,\text{m}$

We need molar conductivity in the form:

$\underline{\hspace{1cm}} \times 10^2\ \text{mS dm}^2\text{ mol}^{-1}$

2. Convert concentration to molarity

A $0.2\%\,(w/v)$ solution means:

$0.2\,\text{g NaOH in }100\,\text{mL solution}$

So in $1\,\text{dm}^3 = 1000\,\text{mL}$, mass of NaOH is:

$0.2 \times \frac{1000}{100} = 2.0\,\text{g}$

Molar mass of NaOH:

$M = 40\,\text{g mol}^{-1}$

Hence molarity:

$c = \frac{2.0}{40} = 0.05\,\text{mol dm}^{-3}$

3. Convert resistivity to conductivity

Given resistivity:

$\rho = 870.0\,\text{m}\Omega\,\text{m} = 870.0 \times 10^{-3}\,\Omega\,\text{m} = 0.870\,\Omega\,\text{m}$

Conductivity is:

$\kappa = \frac{1}{\rho} = \frac{1}{0.870} = 1.1494\,\text{S m}^{-1}$

4. Use molar conductivity formula

Molar conductivity:

$\Lambda_m = \kappa \times \frac{1000}{c}$

when $\kappa$ is in $\text{S cm}^{-1}$ and $c$ in $\text{mol dm}^{-3}$.

Here we have $\kappa$ in $\text{S m}^{-1}$, so better convert directly to $\text{S dm}^2\text{ mol}^{-1}$:

Since

$1\,\text{m}^3 = 1000\,\text{dm}^3$

and $c = 0.05\,\text{mol dm}^{-3} = 50\,\text{mol m}^{-3}$,

$\Lambda_m = \frac{\kappa}{c} = \frac{1.1494}{50} = 0.022988\,\text{S m}^2\text{ mol}^{-1}$

Now convert $\text{S m}^2\text{ mol}^{-1}$ to $\text{mS dm}^2\text{ mol}^{-1}$:

$1\,\text{S m}^2 = 10^4\,\text{mS dm}^2$

Therefore,

$\Lambda_m = 0.022988 \times 10^4 = 229.88\,\text{mS dm}^2\text{ mol}^{-1}$

5. Match required format

$229.88\,\text{mS dm}^2\text{ mol}^{-1} = 2.2988 \times 10^2\,\text{mS dm}^2\text{ mol}^{-1}$

Nearest integer for the blank is:

$\boxed{2}$

6. Comparison with stored answer

Stored correct answer is $23$.

But the computed value is $2.3 \times 10^2$, so the nearest integer to fill in the blank is $2$, not $23$.

It seems the stored answer may have treated the value as $23 \times 10^1$ or missed the formatting of the blank before $\times 10^2$.