Answer
$1.12$ $keV$
Work Step by Step
The maximum kinetic energy ($K_{max}$) of a recoiling electron due to the Compton scattering of an incident beam of energy $E$ can be written as
$K_{max}=\frac{E^2}{E+\frac{mc^2}{2}}$
where, $m$ is the mass of an electron and $c$ is the speed of light in vacuum.
$E=17.5$ $keV$ (given)
$mc^2=\frac{9.1\times 10^{-31}\times (3\times 10^{8})^2}{1.6\times 10^{-19}\times 10^{3}}$ $keV$
or, $mc^2\approx 511.8$ $keV$
$\therefore K_{max}=\frac{(17.5)^2}{17.5+\frac{511.8}{2}}$ $keV$
or, $K_{max}=1.12$ $keV$
Therefore, the maximum kinetic energy of electrons knocked out of a thin copper foil by Compton scattering of an incident beam of 17.5 keV x-rays is $1.12$ $keV$
