using dilution law:
\begin{aligned}
& C _{1} V _{1}= C _{2} V _{2} \\
&2 M \times 400 cm ^{3}=0.2 M \times V _{2} \\
& V _{2}=\frac{2 M \times 400 cm ^{3}}{4000 cm ^{3}}
\end{aligned}
Volume of water to be added \(=V_{2}-V_{1}\)
\begin{aligned}
&=4000 cm ^{3}-400 cm ^{3} \\
&=3600 cm ^{3}
\end{aligned}