A double-slit experiment gives a fringe width of 1.20 mm in air…

Step 1. The index. In the liquid $\beta' = \dfrac{\beta}{n}$. We want $n$, so $n = \dfrac{\beta}{\beta'} = \dfrac{1.20\ \text{mm}}{0.90\ \text{mm}} = 1.33$.

Step 2. The wavelength in the liquid. $\lambda = \dfrac{\lambda_0}{n} = \dfrac{540\ \text{nm}}{1.33} = 405\ \text{nm}$.

Step 3. The new distance. With $\lambda$, $d$ and $\beta = \lambda D/d$ in the liquid, we need $\dfrac{D'}{n} = D$. So $D' = nD = 1.33 \times 1.50\ \text{m} = 2.00\ \text{m}$.