$$ \frac{1}{\lambda_{AC}} = \frac{E_A - E_C}{hc}, \quad \frac{1}{\lambda_{BC}} = \frac{E_B - E_C}{hc} $$

Since the electron transitions from state A to state C in two steps (A to B and then B to C), the energy difference for the transition from A to B is given by

$$ E_A - E_B = (E_A - E_C) - (E_B - E_C) $$

Expressing these energy differences in terms of wavelengths,

$$ \frac{hc}{\lambda_{AB}} = \frac{hc}{\lambda_{AC}} - \frac{hc}{\lambda_{BC}} $$

Cancelling out $hc$, we obtain

$$ \frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}} $$

Substitute the given values:

$$ \frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000} $$

Finding a common denominator:

$$ \frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000} $$

Thus, the wavelength for the A to B transition is

$$ \lambda_{AB} = 3000 \, \mathring{A}. $$

This matches Option D.