$KE$ of colliding bodies before collision $=\large\frac{1}{2} $$mv^2$
After collision the mass $=m+2m=3m$
Velocity becomes $v'=\bigg(\large \frac{m_!-m_2}{m_1+m_2}\bigg)v$
$\qquad= \large\frac{m}{3m} $$v=\large\frac{v}{3}$
KE after collision $=\large\frac{1}{2} $$m \bigg(\large\frac{v}{3}\bigg)^2$
$\qquad=\large\frac{1}{9} \frac{1}{2}$$ mv^2$
$\large\frac{KE _{before}}{KE_{after}}=\large\frac{\Large\frac{1}{2} \large mv^2}{\Large\frac{1}{9} \times \frac{1}{2} \large mv^2}$
$\qquad=9:1$
Hence d is the correct answer.