Answer
Slit width $a=2.99*10^{-3}m$
Work Step by Step
For single slit diffraction, the equation for the minimum is given by
$$asin\theta=m\lambda$$
where $a$ is the slit width, $\theta$ is the angle at which the order minima is formed in the diffraction pattern, $m$ is the order of diffraction, and $\lambda$ is the wavelength of light incident on the slit.
For both incident wavelengths $\lambda1=500nm$ and $\lambda2=600nm$, the slit width $a$ and minima angle $\theta=1.0\ miliradians=0.0573^{\circ}$ remain the same. Thus
$$asin(0.0573)=m\lambda1$$
$$asin(0.0573)=n\lambda2$$
where $m\neq n$.
Since the left hand side of each equation is equal, we can make the right hand side of the equations equal to each other. Hence
$$m(500)=n(600)$$
So that,
$$\frac{m}{n}=\frac{6}{5}$$
Since $m$ and $n$ have to be whole numbers, $m=6$ and $n=5$. We can now use this to find the slit width
$$asin(0.0573)=6(500)$$
Thus,
$$a=2.99*10^{-3}m$$
