Let  p^th bright fringe of first wavelength coincides with  q^th bright fringe of second wavelength
we know,
path difference = $\mathrm{\Delta }x=n\lambda$
Now, path difference for first wavelength  $=p{\lambda }_1$  ----------- (1)
path difference for second wavelength $=q{\lambda }_2$ -------------- (2)
from (1) & (2)

$$\begin{gathered} \Rightarrow p{\lambda }_1=q{\lambda }_2 \\ \Rightarrow \frac{p}{q}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{6000}{4500}=\frac{4}{3} \end{gathered}$$

So,  $p=4$ of first wavelength coincides with $q=3$  of second wavelength.