A body of mass 3kg collides elastically with another body at rest...

Conservation of linear momentum :
$3u = ( 3u / 2 ) + mv ⇒ mv = ( 3u / 2 ) \ldots ( 1 )$

Newton's experimental law :
$u = – e ( u / 2 – v )$

Putting $e = 1$ for elastic collision we obtain,
$v = \dfrac{3u}{2} \ldots . . ( 2 )$
Using (1) and (2), we obtain $m = 1kg$.