A fringe width of 6mm was produced for two slits separated by 1mm...

$\beta = \dfrac{\lambdaD}{d}$

$\lambda = \dfrac{\betad}{D}$

$\lambda = \dfrac{6 \times {10}^{- 3} \times {10}^{- 3}}{10}$

$\lambda = 6 \times {10}^{- 7}m = 600 \times {10}^{- 9}m$

$\lambda = 600nm$