${v^2} = {u^2} + 2as \Rightarrow 0 = {u^2} - 2as \Rightarrow {s_b} = \frac{{{u^2}}}{{2a}}$
Time taken by boggy to stop
$v = u + at \Rightarrow 0 = u - at \Rightarrow t = \frac{u}{a}$
In this time t distance travelled by train$ = {s_t} = ut = \frac{{{u^2}}}{a}$
Hence ratio $\frac{{{S_b}}}{{{S_t}}} = \frac{1}{2}$