Answer
The two astronauts are 20 meters apart.
Work Step by Step
Let $v_1$ be the speed of the lighter astronaut, and let $d_1$ be the distance traveled by this astronaut:
$d_1 = v_1t$
$v_1 = d_1/t$
Let $v_2$ be the speed of the heavier astronaut. We can use conservation of momentum to find an expression for $v_2$:
$m_2~v_2 = m_1~v_1$
$v_2 = \frac{m_1~v_1}{m_2} = \frac{(55~kg)~v_1}{85~kg}$
We can use $v_2$ to find $d_2$:
$d_2 = v_2~t = (\frac{(55~kg)~v_1}{85~kg})(t) = \frac{55~kg}{85~kg}(d_1) = \frac{(55~kg)(12~m)}{85~kg}$
$d_2 = 7.8~m$
The total distance between the astronauts is $d_1 + d_2$ which is 12 m + 7.8 m = 20 meters.
The two astronauts are 20 meters apart.