The angular width of the central maximum in a single slit diffraction pattern is given by $\Delta\theta = \frac{2\lambda}{a}$, which implies that the angular width is directly proportional to the wavelength $\lambda$. Let the initial angular width be $\theta_1 = \theta_0$ for $\lambda_1 = 6000 \text{ \AA}$. When the slit is illuminated by another monochromatic light, the new angular width $\theta_2$ decreases by $30\%$, so $\theta_2 = \theta_0 - 0.30\theta_0 = 0.70\theta_0$. Using the direct proportionality ($\theta \propto \lambda$), we have: $\frac{\theta_2}{\theta_1} = \frac{\lambda_2}{\lambda_1}$ $\frac{0.70\theta_0}{\theta_0} = \frac{\lambda_2}{6000}$ $\lambda_2 = 0.70 \times 6000 \text{ \AA} = 4200 \text{ \AA}$. Thus, the wavelength of this light is $4200 \text{ \AA}$.