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Given:
Frequency of light, $f = 6 \times 10^{14}\text{ Hz}$
Speed of light, $c = 3 \times 10^8\text{ m/s}$
Distance to the screen, $D = 1.55\text{ m}$
Fringe width (distance between centers of two adjacent bright fringes), $\Delta x = 0.75\text{ mm} = 0.75 \times 10^{-3}\text{ m}$

The wavelength of the light is:
$\lambda = \frac{c}{f} = \frac{3 \times 10^8\text{ m/s}}{6 \times 10^{14}\text{ Hz}} = 5 \times 10^{-7}\text{ m}$

The fringe width is given by:
$\Delta x = \frac{\lambda D}{d}$

Therefore, the distance between the slits ($d$) is:
$d = \frac{\lambda D}{\Delta x} = \frac{5 \times 10^{-7}\text{ m} \times 1.55\text{ m}}{0.75 \times 10^{-3}\text{ m}} \approx 1.033 \times 10^{-3}\text{ m} = 1.033\text{ mm}$