Write the reaction

Sodium oxide reacts with water as: $\mathrm{Na_2O + H_2O \rightarrow 2NaOH}$

Calculate molar mass of $\mathrm{Na_2O}$

$M(\mathrm{Na_2O}) = 2(23) + 16 = 62\,\text{g mol}^{-1}$

Calculate moles of $\mathrm{Na_2O}$ used

Given mass = $20.0\,$g

$n(\mathrm{Na_2O}) = \frac{20.0}{62} = 0.3226\,\text{mol}$

Use stoichiometry to find moles of $\mathrm{NaOH}$ formed

From the balanced equation: $1\,\text{mol } \mathrm{Na_2O} \rightarrow 2\,\text{mol } \mathrm{NaOH}$

Therefore, $n(\mathrm{NaOH}) = 2 \times 0.3226 = 0.6452\,\text{mol}$

Calculate molarity of resulting solution

Volume of solution = $500\,$mL = $0.500\,$L

$M = \frac{0.6452}{0.500} = 1.2904\,\text{M}$

Match with the given format

We need: $\text{concentration} = \underline{\hspace{1cm}} \times 10^{-1}\,\text{M}$

Since $1.2904\,\text{M} = 12.904 \times 10^{-1}\,\text{M}$

Nearest integer $= 13$.

Comparison with stored answer

Derived answer = $13$

Stored correct answer = $13$

Hence, the answers agree.