Answer
$49.8\;pm$
Work Step by Step
The $K_{\beta}$ spectral line originates when an electron from the $M$ shell fills a hole in the $K$ shell.
The energy difference between $K$ and $M$ shells of silver is
$\Delta E=(25.51-0.53)\;keV=24.98\;keV$
Therefore, the wavelength of the $K_{\beta}$ line for silver target is
$\lambda_{\beta}=\frac{hc}{\Delta E}=\frac{6.63\times 10^{-34}\times3\times3^{8}}{24.98\times10^3\times1.6\times 10^{-19}}\;m$
$\lambda_{\beta}\approx4.98\times10^{-11}\;m$
$\lambda_{\beta}=49.8\;pm$
