\(400cm^{3} \to 2moldm^{-3}\)

\(0.4 dm^{3} \to 2moldm^{-3}\)

Multiplying in order to get the moles of the acid.

\(0.4dm^{3} \times 2moldm^{-3} = 0.8moles\)

After diluting to \(0.2moldm^{-3}\), the new volume will be

\(\frac{0.8}{0.2} = 4dm^{3}\)

Converting back to cm^3,

We have \(4 \times 1000cm^{3} = 4000cm^{3}\)

Hence, the volume of water added = \((4000 - 400)cm^{3} = 3600cm^{3}\)