For a certain material, the wavelengths of the K_α and L_α X-rays...

The energy of an X-ray photon is related to its wavelength by $E = \dfrac{hc}{\lambda}$.

The transitions corresponding to the given X-rays are:
$K_\alpha$ transition: $L$ shell to $K$ shell ($E_{K\alpha} = E_L - E_K$)
$L_\alpha$ transition: $M$ shell to $L$ shell ($E_{L\alpha} = E_M - E_L$)
$K_\beta$ transition: $M$ shell to $K$ shell ($E_{K\beta} = E_M - E_K$)

From the energy levels:
$E_{K\beta} = E_{K\alpha} + E_{L\alpha}$

Substituting the energy-wavelength relation:
$\dfrac{hc}{\lambda_{K\beta}} = \dfrac{hc}{\lambda_{K\alpha}} + \dfrac{hc}{\lambda_{L\alpha}}$

$\dfrac{1}{\lambda_{K\beta}} = \dfrac{1}{\lambda_{K\alpha}} + \dfrac{1}{\lambda_{L\alpha}}$

Given $\lambda_{K\alpha} = 21.3 \text{ pm}$ and $\lambda_{L\alpha} = 141 \text{ pm}$:
$\dfrac{1}{\lambda_{K\beta}} = \dfrac{1}{21.3} + \dfrac{1}{141}$

$\lambda_{K\beta} = \dfrac{21.3 \times 141}{21.3 + 141}$

$\lambda_{K\beta} = \dfrac{3003.3}{162.3} \approx 18.5 \text{ pm}$