Answer
$56.6\;pm$
Work Step by Step
The $K_{\alpha}$ spectral line originates when an electron from the $L$ shell fills a hole in the $K$ shell.
The energy difference between $K$ and $L$ shells of silver is
$\Delta E=(25.51-3.56)\;keV=21.95\;keV$
Therefore, the wavelength of the $K_{\alpha}$ line for silver target is
$\lambda_{\alpha}=\frac{hc}{\Delta E}=\frac{6.63\times 10^{-34}\times3\times3^{8}}{21.95\times10^3\times1.6\times 10^{-19}}\;m$
$\lambda_{\alpha}=5.66\times10^{-11}\;m$
$\lambda_{\alpha}=56.6\;pm$
