According to the question, the first dark fringe is observed directly opposite to one of the slits. Therefore, half of the fringe width is equal to half of the distance between the slits.

$\frac{\beta }{2}=\frac{d}{2}\Rightarrow \beta =d=6 \mathrm{mm}$

$\Rightarrow \frac{\lambda D}{d}=0.6\times {10}^{-3}\Rightarrow \lambda =\frac{0.6\times {10}^{-3}d}{D}\Rightarrow \lambda =\frac{0.6\times {10}^{-3}\times 0.6\times {10}^{-3}}{80\times {10}^{-2}}$

$\Rightarrow \lambda =450\times {10}^{-9} \mathrm{m}=450 \mathrm{nm}$