Answer
The time between these two events is $~~4.18\times 10^{-7}~s$
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$\gamma = \frac{1}{\sqrt{1-0.900^2}}$
$\gamma = 2.294$
We can find the length of the train in the tunnel view:
$L = \frac{L_0}{\gamma}$
$L = \frac{200~m}{2.294}$
$L = 87.2~m$
In the tunnel view, the length of the tunnel is $200~m$
When the rear of the train reaches the near end of the tunnel (RN), the front of the train has not yet reached the far end of the tunnel (FF).
The front of the train still needs to travel a distance of $112.8~m$
We can find the time this takes:
$t = \frac{112.8~m}{(0.900)(3.0\times 10^8~m/s)}$
$t = 4.18\times 10^{-7}~s$
The time between these two events is $~~4.18\times 10^{-7}~s$
