Answer: Option D
$$\eqalign{
& {\text{Speed of train A}} \cr
& {\text{ = 63 kmph}} \cr
& {\text{ = }}\left( {\frac{{63 \times 5}}{{18}}} \right){\text{m/sec}} \cr
& {\text{ = 17}}{\text{.5 m/sec}} \cr
& {\text{Speed of train B}} \cr
& {\text{ = 54 kmph}} \cr
& {\text{ = }}\left( {\frac{{54 \times 5}}{{18}}} \right){\text{m/sec = 15 m/sec}} \cr
& {\text{If the length of train A be }}x{\text{ metre,}} \cr
& {\text{then}} \cr
& {\text{Speed of train A}} \cr
& {\text{ = }}\frac{{{\text{Length of train + length of platform}}}}{{{\text{Time taken in crossing}}}}{\text{ }} \cr
& \Rightarrow 17.5 = \frac{{x + 199.5}}{{21}} \cr
& \Rightarrow 17.5 \times 21 = x + 199.5 \cr
& \Rightarrow 367.5 = x + 199.5 \cr
& \Rightarrow x = 367.5 - 199.5 \cr
& \Rightarrow 168\,{\text{metres}} \cr
& {\text{Relative speed}} \cr
& {\text{ = ( Speed train A + Speed train B)}} \cr
& {\text{ = (17}}{\text{.5 + 15) m/sec}} \cr
& {\text{ = 32}}{\text{.5 m/sec}} \cr
& {\text{Required time}} \cr
& {\text{ = }}\frac{{{\text{ Length of train A + Length of train B}}}}{{{\text{Relative speed }}}} \cr
& = \left( {\frac{{168 + 157}}{{32.5}}} \right){\text{seconds}} \cr
& = 10\,{\text{seconds}} \cr} $$