A passenger sitting in a train A moving at 90 km /h observes...

Velocity of train A
(\mathrm{V}_{\mathrm{A}}=90 \frac{\mathrm{km}}{\mathrm{hr}}=90 \times \frac{5}{18}=25 \mathrm{ m}/ \mathrm{s})
Velocity of train B
(\mathrm{V}_{\mathrm{B}}=54 \frac{\mathrm{km}}{\mathrm{hr}}=54 \times \frac{5}{18}=15 \mathrm{ m}/ \mathrm{s})
Velocity of train B w.r.t. train
$$\begin{aligned}<br />A=\vec{V}_B & -\vec{V}_A \\<br />& =15-(-25) \mathrm{m}/ \mathrm{s}\\<br />& =40 \mathrm{ m}/ \mathrm{s}<br />\end{aligned}$$
Time of crossing (=\frac{\text{ length of train }}{\text{ relative velocity }})
((8)=\frac{\ell}{40})
(\ell=8 \times40=320) meter.