Relative to the energy of the atom with no vacancy, the energy of a...

The energy of the atom with a vacancy in the $K$, $L$, and $M$ shells are given as $E_K = 25.31 \text{ keV}$, $E_L = 3.56 \text{ keV}$, and $E_M = 0.530 \text{ keV}$.

The energy of the emitted X-ray photon is the difference between the energies of the atom with vacancies in the respective shells.

For the $K_\alpha$ X-ray, the transition is from the $L$ shell to the $K$ shell (the atom goes from a $K$-vacancy to an $L$-vacancy):
$E_{K_\alpha} = E_K - E_L = 25.31 - 3.56 = 21.75 \text{ keV}$

For the $K_\beta$ X-ray, the transition is from the $M$ shell to the $K$ shell (the atom goes from a $K$-vacancy to an $M$-vacancy):
$E_{K_\beta} = E_K - E_M = 25.31 - 0.530 = 24.78 \text{ keV}$

For the $L_\alpha$ X-ray, the transition is from the $M$ shell to the $L$ shell (the atom goes from an $L$-vacancy to an $M$-vacancy):
$E_{L_\alpha} = E_L - E_M = 3.56 - 0.530 = 3.03 \text{ keV}$

Using the relation $\nu = \dfrac{E}{h}$, where $h = 6.63 \times 10^{-34} \text{ J s}$ and $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$:

$\nu_{K_\alpha} = \dfrac{21.75 \times 10^3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 5.25 \times 10^{18} \text{ Hz}$

$\nu_{K_\beta} = \dfrac{24.78 \times 10^3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 5.98 \times 10^{18} \text{ Hz}$

$\nu_{L_\alpha} = \dfrac{3.03 \times 10^3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 7.32 \times 10^{17} \text{ Hz}$