Distance between the slits, $d=0.28 \mathrm{~mm}=0.28 \times 10^{-3} \mathrm{~m}$
Distance between the slits and the screen, $D=1.4 \mathrm{~m}$
Distance between the central fringe and the fourth $(n=4)$ fringe,
$$
u=1.2 \mathrm{~cm}=1.2 \times 10^{-2} \mathrm{~m}
$$

In case of a constructive interference, we have the relation for the distance between the two fringes as:
$$
u=n \lambda \frac{D}{d}
$$

Where,
$$
\begin{aligned}
& n=\text { Order of fringes }=4 \\
& \lambda=\text { Wavelength of light used } \\
& \therefore \lambda=\frac{u d}{n D} \\
& =\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4} \\
& =6 \times 10^{-7} \\
& =600 \mathrm{~nm}
\end{aligned}
$$

Hence, the wavelength of the light is $600 \mathrm{~nm}$.