Now

\({NaOH}: {a} \,{mol}\quad\quad\quad {W}_{{NaOH}}+{W}_{{Na}_{2} {co}_{3}}=4\)

\({Na}_{2} {CO}_{3}: {'} {a}^{\prime} {mol} \quad\quad \Rightarrow 40 {a}+106 {a}=4\)

\(\quad\quad\quad\quad\quad\quad\quad\quad\quad \Rightarrow {a}=\frac{4}{146} \,{~mol}\)

\(\Rightarrow\) therefore mass of \({NaOH}\) is : \(\frac{4}{146} \times 40\, {~g}\) \(=1.095 \approx 1\)