Think in terms of photon energies rather than wavelengths. The $K_\beta$ transition is an electron falling from the $M$ shell straight to the $K$ shell. That drop can be split into two steps: $M \to L$ (which is the $L_\alpha$ line) followed by $L \to K$ (which is the $K_\alpha$ line). Energies add for the combined jump, so $E_{K_\beta} = E_{K_\alpha} + E_{L_\alpha}$.
Because photon energy is inversely proportional to wavelength, the reciprocals add:
\[ \frac{1}{\lambda_{K_\beta}} = \frac{1}{\lambda_{K_\alpha}} + \frac{1}{\lambda_{L_\alpha}}. \]
Numerically this is a resistors-in-parallel style combination:
\[ \lambda_{K_\beta} = \frac{\lambda_{K_\alpha}\,\lambda_{L_\alpha}}{\lambda_{K_\alpha} + \lambda_{L_\alpha}} = \frac{21.3 \times 141}{21.3 + 141} = \frac{3003.3}{162.3}. \]
That gives $\lambda_{K_\beta} \approx 18.5$ pm. The answer must be in picometres, matching option (D).
\[\boxed{\lambda_{K_\beta} \approx 18.5\ \text{pm}}\]