Answer
$\boxed{\lambda_{\beta}=18.5\;pm}$
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 tungsten is
$\Delta E=(69.5-2.30)\;keV=67.2\;keV$
Therefore, the wavelength of the $K_{\beta}$ line for tungsten target is
$\lambda_{\beta}=\frac{hc}{\Delta E}=\frac{6.63\times 10^{-34}\times3\times3^{8}}{67.2\times10^3\times1.6\times 10^{-19}}\;m$
or, $\lambda_{\beta}\approx1.85\times10^{-11}\;m$
or, $\boxed{\lambda_{\beta}=18.5\;pm}$
