Answer
$804\;\rm N$
Work Step by Step
We know, from Newton's second law, that
$$\sum F_y=\dfrac{dp_y}{dt}$$
where we know that the momentum is given by $p_y=mv_y$.
Thus,
$$\sum F_y=\dfrac{d}{dt}(mv_y)$$
where both changes with time, the rocket's mass and velocity.
Thus,
$$\sum F_y=\dfrac{dm}{dt}(v_y)+\dfrac{dv_y}{dt}(m)$$
where $dv_y/dt=a_y$;
$$\sum F_y=\dfrac{dm}{dt}(v_y)+ ma_y $$
Plugging the given;
$$\sum F_y=(-0.5\times120)+ (48\times18)$$
The negative sign in front of the mass is due to the mass losing during its vertical trip.
$$\sum F_y=\color{red}{\bf804}\;\rm N$$