Answer
$\approx 3\;\rm m$
Work Step by Step
To find how far the puck lands from the base of the table, we need to find its velocity at the edge of the table and the net force exerted on it when it is in the air.
The table is frictionless, so the only force exerted on the rocket puck is the thrust force.
$$\sum F_{x_1}=F_T=ma_x$$
where $F_T$ is the thrust force.
Thus,
$$a_x=\dfrac{F_T}{m}=\dfrac{2}{1}=\bf 2\;\rm m/s^2$$
The speed of the rocket at the edge of the table;
$$v_{1x}^2=\overbrace{v_{0x}^2}^{0}+2a_x\Delta x_1$$
$$v_{1x} =\sqrt{2a_x\Delta x_1}=\sqrt{2\times 2\times 4}=\bf 4\;\rm m/s$$
And we know that the front of the rocket is pointed directly toward the edge which means that it is directed toward the positive $x$-direction throughout its trip until it touches the ground.
This means that the thrust is always horizontal and hence the horizontal acceleration component through the whole trip is still 2 m/s/s.
Now we need to find its vertical acceleration component and we know that the only force exerted on it is its own weight due to Earth's gravitational pull.
$$\sum F_y=-mg=ma_y$$
Thus, $$a_y=-g=\bf -9.8\;\rm m/s^2$$
Now we need to find the time of its trip from the edge of the tale to the moment it touches the ground.
$$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$$
At the edge of the table, its vertical velocity component is zero;
$$0=2+(0)t+\frac{1}{2}(-9.8)t^2$$
Hence;
$$t=\sqrt{\dfrac{-2}{-4.9}}=\bf 0.639\;\rm s$$
Therefore, the horizontal distance traveled during this time interval is given by
$$\Delta x_2=v_{1x}t+\frac{1}{2}a_xt^2$$
$$\Delta x_2=(4\times0.639 )+\left(\frac{1}{2}\times 2\times 0.639^2\right)=\color{red}{\bf 2.96}\;\rm m$$
