Answer
$L = 0.111~m$
Work Step by Step
We can find the kinetic energy $K$:
$E = K+E_0$
$K = E - E_0$
$K = 46\times 10^9~eV - 0.511\times 10^6~eV$
$K = 45.999489\times 10^9~eV$
We can find $\gamma$:
$K = (\gamma-1)mc^2$
$\gamma = 1+\frac{K}{mc^2}$
$\gamma = 1+\frac{(45.999489\times 10^9~eV)(1.6\times 10^{-19}~J/eV)}{(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2}$
$\gamma = 89866$
Let $L_0 = 10.0~km$. We can find the length $L$ in the reference frame of the electrons:
$L = \frac{L_0}{\gamma}$
$L = \frac{10000~m}{89866}$
$L = 0.111~m$
