\(\begin{array}{l}
200ml=0,2l\\
n_{SO_3}=\frac{8}{80}=0,1(mol)\\
SO_3+H_2O\to H_2SO_4(1)\\
2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\
Theo\,PT:\,n_{H_2SO_4(1)}=n_{SO_3}=0,1(mol)\\
\to \sum n_{H_2SO_4}=0,2.3+0,1=0,7(mol)\\
Theo\,PT:\,n_{NaOH}=2\sum n_{H_2SO_4}=1,4(mol)\\
\to m_{dd\,NaOH}=\frac{1,4.40}{40\%}=140(g)
\end{array}\)