1. Write the reaction

For complete neutralization:

$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$

So, $1$ mole of $Na_2CO_3$ reacts with $2$ moles of $HCl$.

2. Choose the concordant titre value

Given readings: $4.8, 4.9, 5.0, 5.0, 5.0\ \text{mL}$

By titrimetric convention, we take the concordant readings (same or very close readings). The concordant values are:

$5.0,\ 5.0,\ 5.0\ \text{mL}$

Hence, average titre value:

$V_{HCl} = 5.0\ \text{mL} = 0.005\ \text{L}$

3. Calculate moles of $HCl$ used

Given:

$M_{HCl} = 0.2\ \text{M}$

Moles of $HCl$:

$n_{HCl} = M \times V = 0.2 \times 0.005 = 0.001\ \text{mol}$

4. Find moles of $Na_2CO_3$ in 10.0 mL sample

From stoichiometry:

$1\ \text{mol } Na_2CO_3 \text{ reacts with } 2\ \text{mol } HCl$

Therefore,

$n_{Na_2CO_3} = \frac{0.001}{2} = 0.0005\ \text{mol}$

These moles are present in $10.0$ mL of solution.

5. Calculate molarity of $Na_2CO_3$

Volume of $Na_2CO_3$ solution:

$10.0\ \text{mL} = 0.010\ \text{L}$

So,

$M_{Na_2CO_3} = \frac{0.0005}{0.010} = 0.05\ \text{M}$

6. Convert to mM

$0.05\ \text{M} = 50\ \text{mM}$

7. Final answer

$\boxed{50}$