Answer
$19.68\;keV$
Work Step by Step
The wavelength corresponding to the $K_\beta$ radiation is: $\lambda_\beta=63.0\;pm$
Therefore, the photon energy corresponding to the $K_\beta$ radiation is:
$E_\beta=\frac{hc}{\lambda_\beta}$
or, $E_\beta=\frac{1240\;keV-pm}{\lambda_\beta\;(pm)}$
or, $E_\beta=\frac{1240\;keV-pm}{63.0\;pm}$
or, $E_\beta=19.68\;keV$
