Answer
$9\;\rm m/s$
Work Step by Step
We know, from Newton's second law, that
$$\sum F_x=ma_x$$
and since the force exerted on the 2-kg object is $F_x$,
$$ F_x=ma_x$$
Thus, the acceleration is given by
$$a_x=\dfrac{F_x}{m}=\dfrac{F_x}{2}$$
It is obvious that the force is not constant and hence the acceleration is not constant as well.
This means that we can not use the kinematic equations.
It will be easy to find the acceleration versus time graph in order to make it easy to find the velocity at $t=4$ s.
The velocity then will be the area under the acceleration versus time graph.
Thus, from $t=0$ s, to $t=2$ s, the acceleration is
$$a_x=\dfrac{6}{2}=3\;\rm m/s^2$$
And at $t=4$ s,
$$a_x=\dfrac{0}{2}=0\;\rm m/s^2$$
Now we can draw the graph, as you see below.
$$v_x=\overbrace{v_{ix}}^{0}+\overbrace{a_xt}^{\text{Area under the curve}}$$
$$v_x={\text{Area under the $(a_x-t)$ curve}}\bigg|_0^t$$
$$v_x=A_{\rm yellow-rectangle}+A_{\rm orange-triangle}$$
$$v_x\bigg|_{{\rm at} \;t=4\;\rm s}=(2\cdot 3)+\left(\frac{1}{2}\cdot 2\cdot 3\right)=\color{red}{\bf9}\;\rm m/s$$
