Determine the conductivity from the resistivity.

The resistivity is given as

$$\rho = 870.0\; \text{m}\Omega\cdot\text{m}.$$

Convert this into ohm·meters:

$$870.0\; \text{m}\Omega\cdot\text{m} = 870.0 \times 10^{-3}\; \Omega\cdot\text{m} = 0.87\; \Omega\cdot\text{m}.$$

Conductivity is the reciprocal of resistivity:

$$\kappa = \frac{1}{\rho} = \frac{1}{0.87} \approx 1.15\; \text{S/m}.$$

Find the concentration of NaOH in mol/m³.

A $0.2\%\;(\mathrm{w}/\mathrm{v})$ solution means there are 0.2 grams of NaOH per 100 mL of solution.

In 1 liter (1000 mL) there are:

$$\frac{0.2\; \text{g}}{100\; \text{mL}} \times 1000\; \text{mL} = 2\; \text{g/L}.$$

The molar mass of NaOH is approximately $40\; \text{g/mol}$. Thus, the molarity is:

$$\text{Molarity} = \frac{2\; \text{g/L}}{40\; \text{g/mol}} = 0.05\; \text{mol/L}.$$

Since $1\; \text{L} = 0.001\; \text{m}^3$, converting to SI units (mol/m³):

$$c = 0.05\; \text{mol/L} \times \frac{1}{0.001\; \text{m}^3/\text{L}} = 50\; \text{mol/m}^3.$$

Calculate the molar conductivity in SI units.

Molar conductivity $\Lambda_m$ is given by:

$$\Lambda_m = \frac{\kappa}{c}.$$

Substituting the values:

$$\Lambda_m = \frac{1.15\; \text{S/m}}{50\; \text{mol/m}^3} = 0.023\; \text{S}\cdot \text{m}^2/\text{mol}.$$

Convert the molar conductivity to the desired units (mS dm² mol⁻¹).

First, convert the conductivity:

$$0.023\; \text{S}\cdot \text{m}^2/\text{mol} \times 1000\; \frac{\text{mS}}{\text{S}} = 23\; \text{mS}\cdot \text{m}^2/\text{mol}.$$

Now, convert the area units. Recall that:

$$1\; \text{m}^2 = 100\; \text{dm}^2.$$

So,

$$23\; \text{mS}\cdot \text{m}^2/\text{mol} = 23 \times 100\; \text{mS}\cdot \text{dm}^2/\text{mol} = 2300\; \text{mS}\cdot \text{dm}^2/\text{mol}.$$

The problem asks for the answer in the form

$$\text{[blank]} \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}.$$

Expressing 2300 mS dm²/mol in this form:

$$2300\; \text{mS}\cdot \text{dm}^2/\text{mol} = 23 \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}.$$

Thus, the molar conductivity is

$$\boxed{23 \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}}.$$