Given that the wavelength, thickness of the slab and the shift of the central fringe by the number of fringe widths are, 

$\mathit{\text{λ}}=600 \mathrm{nm}, t=36 \mathrm{\mu m}, n=30$.

If the central bright fringe is shifted by $30\mathit{\text{β}}$, then at the centre point of the screen, ${30}^{\mathrm{th}}$ bright fringe will have a path difference of $30\mathit{\text{λ}}$. Here, $\mathit{\text{β}}=$ Fringe width.

For a bright fringe at the centre of the screen, path difference is, $n\mathit{\text{λ}}=\left(\mathit{\text{μ}}-1\right)t$.

$30\left(600\times {10}^{-9}\right)=\left(\mathit{\text{μ}}-1\right)\left(36\times {10}^{-6}\right)$

$\left(\mathit{\text{μ}}-1\right)=\frac{1}{2} \Rightarrow \mathit{\text{μ}}=\frac{3}{2}=1.5$