Figure shows a circular frictionless track of radius  R, centred at point O. A particle of mass M is  released from point A (OA = R/2). After collision  with the track, the particle moves along the track and in this case , the coefficient of restitution is  $e_1.$. If the coefficient of restitution is $e_2$, particle moves horizontally just after  collision . Then

$\cos \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}$

(a) After collision the particle moves along the track. This means there is no normal component of velocity. Hence  ${\mathrm{e}}_1=\mathrm{o}$
(b) Just before collision, the components of velocity along the normal and along the tangent are  
$\begin{array}{r}{\mathrm{u}}_{\mathrm{n}}=\mathrm{usin}\mathrm{\theta }\\ {\mathrm{u}}_{\mathrm{t}}=\mathrm{ucos}\mathrm{\theta }\end{array}$
 

During collision ${\mathrm{u}}_{\mathrm{t}}$does not change. The normal component of velocity becomes  ${\mathrm{eu}}_{\mathrm{n}}\text{ along }\overrightarrow{\mathrm{B}}\mathrm{O}$ 
 

Question says that velocity of the particle is horizontal after collision, which means 
$$\begin{gathered} {\mathrm{u}}_{\mathrm{t}}\cos \mathrm{\theta }=e_2{u}_{\mathrm{n}}\sin \mathrm{\theta } \\ {\mathrm{ucos}}^2\mathrm{\theta }=e_2{\mathrm{usin}}^2\mathrm{\theta }\Rightarrow {\mathrm{e}}_2={\cot }^2\frac{\mathrm{\pi }}{3}=\frac{1}{3} \end{gathered}$$