Solution:
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr
& \,{\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{by}}\,y\,{\text{m/sec}} \cr
& {\text{Then}},\,\frac{x}{y} = 8\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 8y \cr
& {\text{Now}},\,\frac{{x + 264}}{{20}} = y \cr
& \Rightarrow 8y + 264 = 20y \cr
& \Rightarrow y = 22 \cr
& \therefore {\text{Speed}} = 22\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {22 \times \frac{{18}}{5}} \,{\text{km/hr}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 79.2\,{\text{km/hr}} \cr} $$