$V _{ A }=90 \frac{ km }{ hr }=90 \times \frac{5}{18}=25\,m / s$

Velocity of train $B$

$V _{ B }=54 \frac{ km }{ hr }=54 \times \frac{5}{18}=15\,m / s$

Velocity of train B w.r.t. $\operatorname{train} A=\overrightarrow{ V }_{ B }-\overrightarrow{ V }_{ A }$

$=15-(-25)\,m / s$

$=40\,m / s$

$\text { Time of crossing }=\frac{\text { length of train }}{\text { relative velocity }}$

$(8)=\frac{\ell}{40}$

$\ell=8 \times 40=320 \text { meter. }$