A Young's double slit experiment is performed using a light source...

For the bright fringes of two different wavelengths to coincide on the screen, their distance from the central maximum must be equal.

The position of the $n^{\text{th}}$ bright fringe is given by $y = \dfrac{n\lambda D}{d}$.

Equating the positions for the two wavelengths:
$\dfrac{n_1\lambda_1 D}{d} = \dfrac{n_2\lambda_2 D}{d}$
$n_1\lambda_1 = n_2\lambda_2$

Let $\lambda_1 = 600\text{ nm}$ and $\lambda_2 = 480\text{ nm}$.
$n_1 \times 600 = n_2 \times 480$
$\dfrac{n_1}{n_2} = \dfrac{480}{600} = \dfrac{4}{5}$

The lowest integer values satisfying this ratio are $n_1 = 4$ and $n_2 = 5$.

Therefore, the $4^{\text{th}}$ bright fringe of the $600\text{ nm}$ light coincides with the $5^{\text{th}}$ bright fringe of the $480\text{ nm}$ light.

Answer: $4^{\text{th}}$