Light of a blue laser (wavelength $\lambda=475 \, \text{nm}$) goes through a narrow slit which has width $d$. After the light emerges from the slit, it is visible on a screen that is $ \text {2.013 m} $ away from the slit. The distance between the center of the screen and the first minimum band is $ \text {765 mm} $. Find the width of the slit $d$, in nanometers. 

[i](Proposed by Ahaan Rungta)[/i]
1. **Understanding the problem**: We need to find the width of the slit \( d \) given the wavelength of the laser light \( \lambda = 475 \, \text{nm} \), the distance from the slit to the screen \( L = 2.013 \, \text{m} \), and the distance from the center of the screen to the first minimum \( y = 765 \, \text{mm} \).

2. **Single slit diffraction formula**: The position of the first minimum in a single slit diffraction pattern is given by:
   \[
   d \sin \theta = m \lambda
   \]
   where \( m = 1 \) for the first minimum.

3. **Small angle approximation**: For small angles, \( \sin \theta \approx \tan \theta \approx \theta \). The angle \( \theta \) can be approximated by:
   \[
   \theta \approx \frac{y}{L}
   \]
   where \( y \) is the distance from the center to the first minimum and \( L \) is the distance from the slit to the screen.

4. **Substitute the values**: 
   \[
   \theta \approx \frac{765 \, \text{mm}}{2.013 \, \text{m}} = \frac{765 \times 10^{-3} \, \text{m}}{2.013 \, \text{m}} \approx 0.38
   \]

5. **Using the diffraction formula**:
   \[
   d \sin \theta = \lambda
   \]
   Since \( \sin \theta \approx \theta \):
   \[
   d \cdot 0.38 = 475 \times 10^{-9} \, \text{m}
   \]

6. **Solve for \( d \)**:
   \[
   d = \frac{475 \times 10^{-9} \, \text{m}}{0.38} \approx 1.25 \times 10^{-6} \, \text{m} = 1250 \, \text{nm}
   \]

The final answer is \( \boxed{1250 \, \text{nm}} \).