Answer
$8\;\rm bullets$
Work Step by Step
First, we need to unify the units of speeds of the rocket and the 5-kg bullets.
It is easy to convert one speed rather than 2 of them.
Thus, the speed of one bullet in km/h is
$$v_b=\rm \left(\dfrac{139\;000\;\color{red}{\bf\not} m}{1\;\color{red}{\bf\not} s}\right) \left(\dfrac{1.0\;km}{1000\;\color{red}{\bf\not} m}\right)\left(\dfrac{60^2\;\color{red}{\bf\not} s}{1.0\;h}\right)$$
$$v_b=\bf 500\;400\;\rm km/h$$
This speed isnrelative to the rocket, not the planet.
This means that its speed relative to the planet is
$$v_{fb}=v_{f,rocket}+500\;400\tag 1$$
Now we need to find how many bullets we should fire to reduce the rocket's speed from 25,000 km/h to 15,000 km/h.
The momentum is conserved here, and the bullets are shoted forward to reduce the speed of the rocket because if we shot them backward it will increase its speed.
$$p_i=p_f$$
$$m_{rocket}v_i=(m_{rocket}-Nm_b)v_{f,rocket}+Nm_bv_{fb}$$
where $N$ is the number of fired bullets.
Plug from (1);
$$m_{rocket}v_i=(m_{rocket}-Nm_b)v_{f,rocket}+Nm_b(v_{f,rocket}+500\;400)$$
Solving for $N$;
$$m_{rocket}v_i= m_{rocket}v_{f,rocket}-Nm_b v_{f,rocket}+Nm_b v_{f,rocket}+500\;400Nm_b $$
$$m_{rocket}v_i= m_{rocket}v_{f,rocket} +500\;400Nm_b $$
$$N=\dfrac{m_{rocket}v_i- m_{rocket}v_{f,rocket}}{500\;400 m_b} =\dfrac{m_{rocket}(v_i- v_{f,rocket})}{500\;400 m_b}$$
Plugging the known;
$$N=\dfrac{ 2000 (25000 - 15000)}{500 400 \times 5}\approx \color{red}{\bf 8}\;\rm bullets$$
Therefore, 8 bullets are enough to reduce the rocket's speed as needed.