A block of mass m sliding with speed v on a smooth horizontal...

For a one-dimensional elastic collision, the velocity of the first block after the collision is given by:
$v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}u_1$

Substituting the given values ($m_1 = m$, $m_2 = 2m$, $u_1 = v$):
$v_1 = \dfrac{m - 2m}{m + 2m}v = -\dfrac{v}{3}$

The negative sign indicates that the first block rebounds with a speed of $\dfrac{v}{3}$.

When this block enters the rough patch, the kinetic friction provides a retardation $a = \mu g$.
Using the third equation of motion ($v_f^2 = v_i^2 - 2as$) with final velocity $v_f = 0$:
$0 = \left(\dfrac{v}{3}\right)^2 - 2(\mu g)s$

$s = \dfrac{\dfrac{v^2}{9}}{2\mu g} = \dfrac{v^2}{18\mu g}$

Answer: $\dfrac{v^2}{18\mu g}$