Solution
= 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 $t'_G=\gamma t_G=t'_E$, its reading is
$$
\boxed{t_G=-\frac{vL}{c^2}.}
$$
This is the tunnel clock's reading, not the runner-frame coordinate time $t'_E=-\gamma vL/c^2$. It is also not a calculation of what a photograph of that distant clock would show, which would require an additional light-travel condition.