Given: Distance between the slits, $d = 1\text{ mm} = 10^{-3}\text{ m}$ Distance of the screen, $D = 1\text{ m}$ Wavelength of monochromatic light, $\lambda = 500\text{ nm} = 500 \times 10^{-9}\text{ m}$

The width of the central maximum in a single-slit diffraction pattern is given by $W_c = \frac{2\lambda D}{a}$, where $a$ is the width of each slit. The fringe width in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.

According to the question, the central maximum of the single-slit pattern contains $10$ maxima of the double-slit pattern. Therefore, $W_c = 10 \beta$ $\frac{2\lambda D}{a} = 10 \left(\frac{\lambda D}{d}\right)$ $\frac{2}{a} = \frac{10}{d}$ $\implies a = \frac{2d}{10} = \frac{d}{5}$

Substituting the value of $d = 1\text{ mm}$: $a = \frac{1\text{ mm}}{5} = 0.2\text{ mm}$.