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 : $\frac{4}{146} \times 40\, {~g}$ $=1.095 \approx 1$