C

Step 1: The condition for the first diffraction minimum for a single slit is given by the formula: \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \lambda \) is the wavelength of the light, and \( m \) is the order of the minimum (1, 2, etc.).
Step 2: Given:
- Wavelength, \( \lambda = 6000 \times 10^{-10} m \)
- Slit width, \( a = 0.6 \times 10^{-4} m = 6.0 \times 10^{-5} m \)
Step 3: For maxima to be formed on both sides of the central maximum, the maximum value of \( m \) is determined by the condition \( \sin \theta \leq 1 \):
\( m \lambda \leq a \)
\( m \leq \frac{a}{\lambda} = \frac{6.0 \times 10^{-5}}{6000 \times 10^{-10}} = 10 \)
Therefore, the maximum possible number of minima produced on both sides of the central maximum is \( 10 + 10 = 20 \).
Thus, the answer is 20, which corresponds to option C.