Answer
$5.51\;\rm m/s^2$
Work Step by Step
At the moment the bolt is separated from the rocket, it must have the same speed as the rocket. So, it will move up for a while until its vertical velocity component became zero then it starts to fall back again toward the ground.
Do not be confused by this situation, it is similar to if you fired the bolt upward by a velocity that is equal to the rocket after 4 seconds from lift-off.
We know that the bolt journey takes 6 s, but we need to find its initial velocity which is as same as the rocket's velocity.
$$v_y=v_{iy}+a_yt$$
The rocket accelerates from rest, so
$$v_y=0+a_yt=4a_y$$
$$v_y= 4a_y\tag 1$$
Thus the velocity of the rocket after 4 seconds and the velocity of the bolt- when it is separated from the rocket- are equal and they are given by (1).
Now the bolt will have two stages of motion, the first upward until its vertical velocity component became zero and the second the falling toward the ground.
Both stages are under free-fall acceleration and we will ignore any air resistance.
Now we need to find the initial height of the bolt at the moment it separated from the rocket (which is the height of the rocket as well after 4 seconds from left-off)
$$y=\overbrace{ y_i}^{0} +\overbrace{ v_{iy}t}^{0} +\frac{1}{2}a_yt^2$$
The rocket accelerates from rest and its initial height was the ground.
$$y= \frac{1}{2}a_yt^2=\frac{1}{2}a_y\cdot 4^2$$
$$y=8a_y\tag 2$$
Now we have the initial velocity of the bolt and its initial height.
We also know that its final height will be the ground at which $y=0$.
Noting that the whole trip of the bolt is under the free-fall.
$$\overbrace{ y_b }^{0} =y+v_{y}t-\frac{1}{2}gt^2$$
Plugging from (1) and (2);
$$0 =8a_y+4a_yt-\frac{1}{2}gt^2=8a_y+(4a_y\cdot 6)-\left(\frac{1}{2}g\cdot 6^2\right)$$
Thu,
$$a_y=\dfrac{\frac{36}{2}\cdot 9.8}{32}=\color{red}{\bf 5.51}\;\rm m/s^2$$
