\(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{Na2O}=\frac{6,2}{62}=0,1\left(mol\right)\)
\(\Rightarrow n_{NaOH}=n_{Na2O}=0,1\left(mol\right)\)
\(m_{NaOH_{pư}}=m.10\%+0,2.40=0,1m+8\left(g\right)\)
\(m_{dd}=m+6,2\left(g\right)\)
\(\frac{0,1m+8}{m+6,2}.100\%=20\%\)
\(\Rightarrow m=67,6\left(g\right)\)