 Solution

The entrance clock is at $x=0$. Let $G$ be the event on that clock simultaneous with emission in the runner's frame. Since $t_G^{\prime }=\gamma t_G=t_E^{\prime }$, its reading isThis is the tunnel clock's reading, not the runner-frame coordinate time $t_E^{\prime }=-\gamma vL/c^2$. It is also not a calculation of what a photograph of that distant clock would show, which would require an additional light-travel condition.