Answer
$ 6.03 \;\rm cm$
Work Step by Step
Our system undergoes a damping simple harmonic motion, so the maximum displacement is given by
$$x_{(t)}=Ae^{-t/2\tau}$$
And since our system has an amplitude that decreases by 2.0% during each complete oscillation, so $x_{(t)}=0.98 A$.
$$ 0.98\color{red}{\bf\not} A=\color{red}{\bf\not} Ae^{-t/2\tau}$$
Thus,
$$\dfrac{-t}{2\tau}=\ln(0.98)$$
Now we can find $\tau$ since we know the period of one oscillation which is 0.5 s
$$\tau=\dfrac{-T}{2\ln(0.98)}\tag 1$$
Now we know that the time our system would take for completing 25 full oscillations is $t=25T$ and hence its amplitude is given by
$$x_{(25T)}=Ae^{-t/2\tau}=Ae^{-25T/2\tau}$$
Plugging from (1);
$$x_{(25T)}=Ae^{-\color{red}{\bf\not} 2(25\color{red}{\bf\not} T)\ln(0.98)/-\color{red}{\bf\not} 2\color{red}{\bf\not} T}=Ae^{25T\ln(0.98)/ T}$$
$$x_{(25T)}= Ae^{ 25 \ln(0.98) }$$
We know that the initial amplitude is 10 cm, so
$$x_{(25T)}= 10e^{ 25 \ln(0.98) }$$
$$x_{(25T)}=\color{red}{\bf 6.03}\;\rm cm$$
