During the transition of electron from state A to state C of a Bohr atom,  the wavelength of emitted radiation is  2000 A˚  and it becomes  6000 A˚  when the electron jumps from state B to state C.  Then the wavelength of the radiation emitted during the transition of electrons  from state A to state B is:

Transition A→C corresponds to $$\lambda_1 = 2000$$ Å and B→C to $$\lambda_2 = 6000$$ Å. We denote the wavelength for A→B by $$\lambda_3$$. Since the energy of A→C equals the sum of energies of A→B and B→C, $$E_{A→C} = E_{A→B} + E_{B→C}$$. Expressing each energy as $$hc/\lambda$$ gives $$\frac{hc}{\lambda_1} = \frac{hc}{\lambda_3} + \frac{hc}{\lambda_2}$$. Substituting the given wavelengths yields $$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{2000} - \frac{1}{6000} = \frac{3-1}{6000} = \frac{1}{3000}$$.

This gives $$\lambda_3 = 3000$$ Å. Therefore the correct answer is $$3000$$ Å.