Answer
${\bf 0.41}\;\rm m/s$
Work Step by Step
We need to find the force exerted by the radiation pressure.
We know that the radiation pressure on an object that absorbs all the radiation is given by
$$P=\dfrac{I}{c}\tag 1$$
where $P$ here is for pressure.
The force exerted by this radiation is then given by
$$F=PA$$
where $P$ is the pressure and $A$ is the cross-sectional area of the object.
Plug from (1),
$$F=\dfrac{I}{c}A$$
where $I={\rm Power}/A$,
$$F=\dfrac{ { P\rm ower}}{cA}A$$
$$F=\dfrac{ {P\rm ower}}{c } \tag 2$$
Now we need to use Newton's second law, knowing that the only force exerted on the block is the light force (radiation force).
Thus,
$$F=ma$$
Plug from (2),
$$\dfrac{ {P\rm ower}}{c }= ma$$
Solving for $a$,
$$a=\dfrac{ {P\rm ower}}{m \;c}\tag 3$$
Now we need to find the final velocity of the block after it moves 100 m starting from rest.
$$v_f^2=v_i^2+2a\Delta x=0+2a\Delta x$$
So,
$$v_f=\sqrt{2a\Delta x}$$
Pluf from (3),
$$a= $$
$$v_f=\sqrt{2 \Delta x\dfrac{ {P\rm ower}}{m \;c}}$$
Plug the known;
$$v_f=\sqrt{2 (100)\dfrac{ (25\times 10^6)}{(100)(3\times 10^8)}}$$
$$v_f=\color{red}{\bf 0.408}\;\rm m/s$$
