From the interference formula $sin\theta =\frac{m\lambda }{d }$ , considering that at first bright fringe m = 3, we obtain $d=\frac{\lambda }{sin\theta }$

We find, considering the position of the first fringe with respect to the central fringe, :

$tg\theta =\frac{▵y}{L}=\frac{5.36}{875}=\mathrm{0.61257...}\to sin\theta =sin\left(arctg\left(\mathrm{0.61257...}\right)\right)\simeq 0.61256$

Finally, we obtain:     $d=\frac{\lambda }{sin\theta }=\frac{546\cdot 10^{-9}}{0.61256}\simeq 89.1\cdot 10^{-6}=89.1\mu m$