Answer
$402\;\rm m$
Work Step by Step
We can use the conservation of the momentum principle since the system's momentum is conserved in the explosion.
We assume that the two fragments are moving in the $x$-direction.
Thus,
$$p_{ix}=p_{fx}$$
The system [the object as one piece] before the explosion was at rest which means that the initial momentum is zero.
Hence,
$$p_{fx}=m_1v_{fx,1}+m_2v_{fx,2}=0$$
where $m_1$ is the heavier fragment. Thus, $m_1=7m_2$, and hence,
$$7\color{red}{\bf\not} m_2 v_{fx,1}+\color{red}{\bf\not} m_2v_{fx,2}=0$$
$$v_{fx,2} =-7 v_{fx,1}\tag 1 $$
where $v_{fx,1}$ and $v_{fx,2}$ are the initial velocities of the two fragments after the collision to the final stop.
Let's assume that $m_1$ moves to the right while $m_2$ moves to the left.
Now we need to apply Newton's second law on the heavier fragment to find the coefficient of friction. See the figures below.
$$\sum F_{y1}=F_{n1}-7mg=7ma_y=7m(0)=0$$
Thus,
$$F_{n1}=7mg\tag 2$$
$$\sum F_{x1}=-f_{k1}=-\mu_k F_{n1}=7ma_x$$
Plugging from (2);
$$-\mu_k \color{red}{\bf\not}7\color{red}{\bf\not} mg=\color{red}{\bf\not}7\color{red}{\bf\not} ma_x$$
Hence,
$$a_x=-\mu_k g\tag3 $$
Now we can use the kinematic formula of
$${v_{fx,1}^{2'}}=v_{fx,1}^2+2a_x\Delta x_1$$
where the initial velocity here is the final velocity just after the collision and the final velocity here is zero since it finally stops.
$$0=v_{fx,1}^2+2a_x\Delta x_1$$
Plugging from (3);
$$ v_{fx,1}^2=2\mu_k g\Delta x_1$$
$$ v_{fx,1} =\sqrt{2\mu_k g\Delta x_1}$$
Plugging into (1);
$$v_{fx,2} =-7 \sqrt{2\mu_k g\Delta x_1}\tag 4$$
Now we need to apply Newton's second law on the second fragment to find its acceleration. We can find that we will get the same acceleration as the first one since both are having the same coefficient of friction.
Now we can use the kinematic formula of
$${v_{fx,2}^{2'}}=v_{fx,2}^2+2a_x\Delta x_2$$
where the initial velocity here is the final velocity just after the collision and the final velocity here is zero since it finally stops.
$$0=v_{fx,2}^2+2a_x\Delta x_2$$
$$-v_{fx,2}^2=2a_x\Delta x_2$$
Plugging from (3) and (4);
$$-(-7 \sqrt{2\mu_k g\Delta x_1}) ^2=-2\mu_k g \Delta x_2$$
$$ 98 \mu_k g\Delta x_1 =-2\mu_k g \Delta x_2$$
Solving for $\Delta x_2$;
$$\Delta x_2=\dfrac{98 \color{red}{\bf\not} \mu_k \color{red}{\bf\not} g\Delta x_1 }{-2\color{red}{\bf\not} \mu_k \color{red}{\bf\not} g }$$
$$\Delta x_2=- 49 \Delta x_1=-49\times 8.2 $$
$$\Delta x_2\approx- \color{red}{\bf 402}\;\rm m $$
the negative sign is due to direction.
