Answer
$1.23\times 10^3\;\rm m/s$
Work Step by Step
The momentum is conserved through the explosion, so
$$p_{ix}=p_{fx}$$
$$(m_1+m_2)v_{ix}=m_1v_{fx,1}+m_2v_{fx,2}$$
We know that the two stages will separate after the explosion, and we also know that the first stage is 3 times the mass of the first stage. So,
$m_1=3m_2$ and hence, $m_1=3m$ and $m_2=m$. See the figure below.
$$(3m+m)v_{ix}=3m v_{fx,1}+mv_{fx,2}$$
$$4\color{red}{\bf\not} mv_{ix}=3\color{red}{\bf\not} m v_{fx,1}+\color{red}{\bf\not} mv_{fx,2}$$
$$4 v_{ix}=3 v_{fx,1}+ v_{fx,2}\tag 1$$
We are given the final velocity of the first stage backward relative to the first stage.
Thus, its velocity relative to the ground is given by
$$v_{fx,1}=v_{fx,2}-35$$
Plugging into (1);
$$4 v_{ix}=3 (v_{fx,2}-35)+ v_{fx,2} $$
$$4 v_{ix}=3 v_{fx,2}-105+ v_{fx,2} $$
$$4 v_{ix}=4v_{fx,2}-105 $$
Solving for $v_{fx,2}$;
$$ v_{fx,2}=\dfrac{4 v_{ix}+105}{35} =\dfrac{(4 \times 1200)+105}{4} $$
$$ v_{fx,2}=\color{red}{\bf 1.23\times 10^3}\;\rm m/s $$
