Answer
$\approx 4.5\;\rm km$
Work Step by Step
First, we need to find the highest point just before the explosion occurs. Then we need to find the time the trip would take for the heaviest fragment to fall directly down, so we can find the horizontal distance traveled by the other fragment.
Using the kinematic formula of velocity squared to find the highest point
$$\overbrace{v_{1y}^2}^{0}=(\overbrace{v_{i y}}^{v_0\sin\theta})^2+2 \overbrace{a_y}^{-g}(y_{max}-\overbrace{y_i}^{0})$$
Thus,
$$0=(v_0\sin\theta)^2-2gy_{max}$$
Thus,
$$y_{max}=\dfrac{v_0^2\sin^2\theta}{2g}=\dfrac{125^2\sin^255^\circ}{2\times 9.8}$$
$$y_{max}=\bf 535\;\rm m\tag 1$$
Now we need to find the time interval of this stage of motion from $y=0$ to $y_{max}$.
$$v_{fy}=v_{iy}+a_yt=v_0\sin\theta-gt$$
where $v_{fy}$ at the highest point is zero. Thus,
$$0=v_0\sin\theta-gt$$
$$t=\dfrac{v_0\sin\theta}{g}=\dfrac{125\sin55^\circ}{9.8}=\bf 10.45\;\rm s$$
Hence, the horizontal distance traveled during this time is given by
$$\Delta x_1=v_xt=v_0\cos\theta t=125\cos 55^\circ\times 10.45=\color{blue}{\bf749 }\;\rm m$$
Now we need to find the horizontal velocity component of the lighter fragment at the highest point. The momentum is conserved during the explosion, so that
$$p_{ix}=p_{fx}$$
$$(m_L+m_H)v_{ix}=m_Lv_{xL,1}$$
where $L$ is for lighter and $H$ for heavier. Recall that $m_L=m$, and hence, $m_H=4m$.
Hence,
$$(m+4m)v_{0}\cos\theta=m v_{xL,1}$$
$$5\color{red}{\bf\not}
m v_{0}\cos\theta=\color{red}{\bf\not}
m v_{xL,1}$$
$$v_{xL,1}=5\times 125\cos 55^\circ=\bf 358.5\;\rm m/s$$
The time this stage would take is the same time as the first stage.
Thus,
$$\Delta x_2=v_{xL,1}t_2 =358.5\times 10.45=\color{blue}{\bf3746 }\;\rm m$$
Therefore, the total distance is given by
$$\Delta x_{tot}=\Delta x_1+\Delta x_2=749+3746 $$
$$\Delta x_{tot}= \bf 4495\;\rm m\approx \color{red}{\bf 4.5}\;\rm km$$
