In a double-slit experiment the slits are 0.36 mm apart and the screen is 1.50 m away…

Step 1. The fringe width. Between the 4th bright fringe on the left and the 4th on the right there are $8$ fringe widths. So $\beta = \dfrac{16.0\ \text{mm}}{8} = 2.00\ \text{mm}$.

Step 2. The wavelength. We want $\lambda$, so $\lambda = \dfrac{\beta d}{D} = \dfrac{2.00 \times 10^{-3}\ \text{m} \times 0.36 \times 10^{-3}\ \text{m}}{1.50\ \text{m}} = 480\ \text{nm}$.

Step 3. The 2nd dark fringe. The first dark fringe is at $\tfrac12\beta$, so the second is at $\tfrac32\beta = 3.00\ \text{mm}$.

Step 4. The new screen. $\beta$ is in proportion to $D$. So $\beta' = 2.00\ \text{mm} \times \dfrac{1.80\ \text{m}}{1.50\ \text{m}} = 2.40\ \text{mm}$.