Body A of mass 4m moving with speed u collides with another body B...

$4mu + 0 = 4m{v}_{1} + 2m{v}_{2}$
$⇒ 2u = 2{v}_{1} + {v}_{2} \ldots ( A )$
For elastic collision,
$e = 1$
$⇒ u = {v}_{2} - {v}_{1} \ldots ( B )$

From \left(A\right)$and \left(B\right)$,
$4u = 3{v}_{2} ⇒ {v}_{2} = \dfrac{4}{3}u$
and ${v}_{1} = \dfrac{u}{3}$
Fraction of energy lost by $( A )$,
$= \dfrac{\dfrac{1}{2}4m{u}^{2} - \dfrac{1}{2}4m{v}_{1}^{2}}{\dfrac{1}{2}4m{u}^{2}}$
$= \dfrac{{u}^{2} - \dfrac{{u}^{2}}{9}}{{u}^{2}} = \dfrac{8}{9}$