The width of fringe is 2mm on the screen in a double slit...

For a light of wavelength $\lambda$, the fringe width in double slit experiment is given by

$\beta = \dfrac{\lambdaD}{d} . . . ( 1 )$

For a light of wavelength $\lambda '$, the fringe width is given by

$\beta ' = \dfrac{\lambda ' D}{d} . . . \left(2\right)$

Divide equation (2) by equation (1) and simplify to obtain the fringe width for the later case.

$\begin{cases}\dfrac{\beta '}{\beta} & = & \dfrac{\dfrac{\lambda ' D}{d}}{\dfrac{\lambdaD}{d}} \\ & = & \dfrac{\lambda '}{\lambda} \\ & ⇒ & \beta ' = \dfrac{\lambda '}{\lambda}\beta . . . \left(3\right)\end{cases}$

Substitute the values of the known parameters into equation (3) to calculate the required fringe width.

$\begin{cases}\beta ' & = & \dfrac{600nm}{400nm} \times 2mm \\ & = & 3mm\end{cases}$