A student lights a single slit with light of wavelength 600 nm…

Step 1. The slit width. The two first dark places are the ends of the central maximum, so $w = 7.2$ mm. We want $a$, so $a = \dfrac{2\lambda D}{w} = \dfrac{2 \times 600 \times 10^{-9}\ \text{m} \times 1.2\ \text{m}}{7.2 \times 10^{-3}\ \text{m}} = 0.200\ \text{mm}$.

Step 2. The second dark place. The first is at $\dfrac{w}{2} = 3.6$ mm. The second is at $2 \times 3.6\ \text{mm} = 7.20\ \text{mm}$.

Step 3. The new slit. $w' = \dfrac{2\lambda D}{a'} = \dfrac{2 \times 600 \times 10^{-9}\ \text{m} \times 1.2\ \text{m}}{0.40 \times 10^{-3}\ \text{m}} = 3.60\ \text{mm}$.