Fringe width, $\beta =\frac{\lambda \mathrm{D}}{\mathrm{d}} ...\left(i\right)$
where, D: distance between screen and slit
d: distance between two slits
when a film of thickness t and refractive index $\mu$ is placed over one of the slit, the fringe pattern is shift by distance S and is given by $\mathrm{S}=\frac{(\mu -1)\mathrm{tD}}{\mathrm{d}} ...\left(ii\right)$

Given: $S=10 \beta ...\left(iii\right)$
From equations (i), (ii) and (iii), we get

$\frac{(\mu -1)+\mathrm{D}}{\mathrm{d}}=10\frac{\lambda \mathrm{D}}{\mathrm{d}}$

 $\Rightarrow \mu -1=\frac{10\lambda }{t}=\frac{10\times 6000\times {10}^{-9}\mathrm{cm}}{3\times {10}^{-3}\mathrm{cm}}$

$\Rightarrow \mu -1=0.2$

$\Rightarrow \mu =1.2$