Answer
$\theta = 92.6^{\circ}$
Work Step by Step
We can express $85~km/h$ in units of $m/s$:
$v_1 = (85~km/h)(\frac{1~h}{3600~s})(\frac{1000~m}{1~km}) = 23.61~m/s$
Suppose the bullet is fired from an angle $\theta$ with the track, where $\theta$ is measured from the train's direction of motion.
Then after entering the train car, the component of the bullet's velocity parallel to the track must be equal to the train's velocity.
We can find $\theta$:
$(0.80)~v_2~cos~(180^{\circ}-\theta) = v_1$
$cos~(180^{\circ}-\theta) = \frac{v_1}{(0.80)~v_2}$
$180^{\circ}-\theta = cos^{-1}~(\frac{v_1}{(0.80)~v_2}$
$180^{\circ}-\theta= cos^{-1}~[\frac{23.61~m/s}{(0.80)~(650~m/s)}]$
$180^{\circ}-\theta = 87.4^{\circ}$
$\theta = 180^{\circ}-87.4^{\circ}$
$\theta = 92.6^{\circ}$
Note that it is not necessary to know the width of the train car since the component of the bullet's velocity along the track is equal to the train's velocity.
