To find the point of coincidence of bright fringes, we can equate the distance of bright fringes from the central maximum, made by both the wavelengths of light. 

Given, distance between the screen and slit,
D = 120 cm = 120 x 10^-2 m
Slit width, d = 2 mm = 2 x 10^-3 m
$\lambda$₁ = 6500 $\stackrel{0}{\mathrm{A}}$ and $\lambda$₂ = 5200 $\stackrel{0}{\mathrm{A}}$
Let nth fringe due to  $\lambda$₂ = 5200 $\stackrel{0}{\mathrm{A}}$ coincide with (n $-$1) th
bright fringe due to  $\lambda$₁ = 6500 $\stackrel{0}{\mathrm{A}}$.

$\therefore \frac{Dn{\lambda }_2}{d}=\frac{D(n-1){\lambda }_1}{d}$

or    n  x  5200 = (n $-$ 1) 6500
or   4n = 5n $-$ 5

$\Rightarrow$  n = 5
So,  the least distance required,  $y_n=\frac{n{\lambda }_{2}D}{d}$

$=\frac{5\times 5200\times {10}^{-10}\times 120\times {10}^{-2}}{2\times {10}^{-3}}$

$=0.16\times {10}^{-2} \mathrm{m}=0.16 \mathrm{cm}$