From the figure geometry: the composite plate in front of each slit has total width 12 $\mu$m (the wedge thickness). In front of the upper slit, wedge $A$ has thickness $x$ (over 1 mm height) and wedge $B$ has thickness $y$, with $x+y=12\mu$m. The slant geometry gives, using similar triangles with the 1 mm marker over the 4 mm vertical extent shown: $x=3\mu$m and $y=9\mu$m for the upper slit. By symmetry, the lower slit sees $x=9\mu$m of $A$ and $y=3\mu$m of $B$.

Extra optical path introduced at each slit equals $(n-1)\times (\text{thickness})$. The path-length difference $\Delta$ introduced by the plates (upper minus lower path) is

$\Delta =({\mu }_A-1)(3)+({\mu }_B-1)(9)-({\mu }_A-1)(9)-({\mu }_B-1)(3)$

$=({\mu }_A-1)(3-9)+({\mu }_B-1)(9-3)=-6({\mu }_A-1)+6({\mu }_B-1)=6({\mu }_B-{\mu }_A)$

With ${\mu }_A=1.7$ and ${\mu }_B=1.5$: $\Delta =6(1.5-1.7)=-1.2\mu$m.

The central maximum shifts by $y_{\text{shift}}=\frac{\Delta \cdot D}{d}$ (magnitude):

$y_{\text{shift}}=\frac{1.2\times 10^{-6}\times 2}{2\times 10^{-3}}=1.2\times 10^{-3}$ m $=1.20$ mm.