Answer
$7.57\;\rm cm$
Work Step by Step
Let's say that Fred and Brutus are met head-on at the origin at which $x=0$.
The momentum is conserved (from just before the collision to just after it) so that
$$p_{ix}=p_{fx}$$
$$m_Fv_{ix,F}+m_Bv_{ix,B}=p_{fx}$$
The two of them after the collision will move as one unit; thus
$$m_Fv_{ix,F}+m_Bv_{ix,B}=(m_F+m_B)v_{fx}$$
where $v_{fx}$ is their velocity just after the collision.
We assume that Fred is moving to the right, which is the positive direction, and Brutus is moving to the left.
Solving for $v_{fx}$;
$$v_{fx}=\dfrac{m_Fv_{ix,F}+m_Bv_{ix,B}}{(m_F+m_B)}$$
Plugging the known;
$$v_{fx}=\dfrac{(60\times 6)+(120\times- 4)}{(60+120)}=\bf -0.667\;\rm m/s$$
and since the final speed is negative, they will move to the left which was Brutus's initial direction.
Now we need to find the distance traveled by them as one unit while they are sliding on the ground.
$$\overbrace{v_{fx,1} ^2}^{0}=v_{fx}^2+2a_x d$$
Hence,
$$d=\dfrac{-v_{fx}^2}{2a_x}=\dfrac{-(-0.667)^2}{2a_x}\tag 1$$
Now we need to find the acceleration. We know that there are two forces exerted on them vertically which are their own weight and the normal force. We also know that there is one force exerted on them horizontally which is the kinetic friction force toward the right since they are sliding toward the left.
$$\sum F_x=f_s=-(m_F+m_B)a_x$$
where $f_s=\mu_kF_n$;
$$\mu_kF_n=-(m_F+m_B)a_x\tag 2$$
$$\sum F_y=F_n-(m_F+m_B)g=(m_F+m_B)a_y=(m_F+m_B)(0)=0$$
Thus,
$$F_n=(m_F+m_B)g$$
Plugging into (2);
$$\mu_k(m_F+m_B)g=-(m_F+m_B)a_x $$
So,
$$-\mu_k g= a_x $$
Plugging into (1);
$$d =\dfrac{-(-0.667)^2}{2(-\mu_k g) } =\dfrac{ (-0.667)^2}{2( 0.3\times 9.8 ) } $$
$$d=0.0757\;\rm m\approx \color{red}{\bf7.57}\;cm$$
