${{n}_{N{{a}_{2}}O}}=\frac{{{m}_{N{{a}_{2}}O}}}{{{M}_{N{{a}_{2}}O}}}=\frac{12,4}{62}=0,2\left( mol \right)$

$N{{a}_{2}}O+{{H}_{2}}O\to 2NaOH$

0,2               $\to $  0,4             (mol)

Theo phương trình :

${{n}_{NaOH}}=2.{{n}_{N{{a}_{2}}O}}=2.0,2=0,4$ (mol)

   $\Rightarrow $${{C}_{{{M}_{NaOH}}}}=\frac{{{n}_{NaOH}}}{V}=\frac{0,4}{2}=0,2\left( M \right)$

$\Rightarrow $Chọn đáp án B