Answer
$99\;\rm m$
Work Step by Step
We know, from Newton's third law, that the force exerted by the Federation starship on the shuttlecraft is equal to the force exerted by the shuttlecraft on the Federation starship in magnitude but opposite in direction.
And since there is no friction in space, the two objects will move toward each other at a constant acceleration.
To understand the situation perfectly, see the figure below.
We chose the origin point to be the initial position of the starship where the shuttlecraft is 10 km away from this origin.
Now we know that the initial velocities of the two objects are zeros and we also know that the net force exerted on both of them is the same.
We need to apply Newton's second law to find the acceleration;
$$\sum F_{1\;on\;2}=\sum F_{2\;on\;1}=F_{beam}$$
where $1$ is the shuttlecraft and $2$ is the starship.
Thus the acceleration of the starship is given by
$$F_{beam}=m_{2}a_2$$
$$a_2=\dfrac{F_{beam}}{m_2}$$
Plugging the known;
$$a_2=\dfrac{4\times 10^4 }{2\times 10^6}=\bf 2\times 10^{-2}\;\rm m/s^2$$
And the acceleration of the shuttlecraft is given by
$$a_1=\dfrac{-F_{beam}}{m_1}$$
The negative sign is due to the direction, see the figure below.
Plugging the known;
$$a_1=\dfrac{-4\times 10^4 }{2\times 10^4}=\bf -2 \;\rm m/s^2$$
We know that both objects will meet at some point of $x_1=x_2=x$, after a time interval of $t_1=t_2=t$.
Thus;
$$x_2=\overbrace{x_{i2}}^{0}+\overbrace{v_{ix,2}t_2}^{0}+\frac{1}{2}a_2t_2^2$$
So that
$$x = \frac{1}{2}a_2t^2\tag 1$$
By the same approach;
$$x_1= x_{i1} +\overbrace{v_{ix,1}t_1}^{0}+\frac{1}{2}a_1t_1^2$$
$$x =x_{i1}+\frac{1}{2}a_1t ^2$$
Plugging into (1);
$$x_{i1}+\frac{1}{2}a_1t ^2= \frac{1}{2}a_2t^2 $$
Solving for $t$ and then plugging the known;
$$x_{i1}= \frac{1}{2}a_2t^2 -\frac{1}{2}a_1t ^2$$
$$x_{i1}= \left[ \frac{1}{2}a_2 -\frac{1}{2}a_1\right]t^2$$
$$\dfrac{x_{i1}}{ \left[ \frac{1}{2}a_2 -\frac{1}{2}a_1\right]}=t^2$$
$$t=\sqrt{\dfrac{x_{i1}}{ \left[ \frac{1}{2}a_2 -\frac{1}{2}a_1\right]}}$$
The initial position of the shuttlecraft is 10 km which is 10 000 m.
$$t=\sqrt{\dfrac{10000}{ \left[ \frac{ 2\times 10^{-2}}{2}-\frac{-2}{2} \right]}}$$
$$t=\bf 99.5 \;\rm s$$
Plugging into (2);
$$x = \frac{1}{2} \times 2\times 10^{-2}\times 99.5^2$$
$$x =\color{red}{\bf 99.0}\;\rm m$$
