Answer
$\approx 25\;\rm s$
Work Step by Step
Our system undergoes a damping simple harmonic motion, so the maximum displacement is given by
$$x_{(t)}=Ae^{−t/2\tau}$$
The ball amplitude became half the initial amplitude after 30 oscillations which means after $t=30T$, so
$$0.5\color{red}{\bf\not} A=\color{red}{\bf\not} Ae^{−30T/2\tau}$$
Thus,
$$\dfrac{−30T}{2\tau}=\ln(0.5)$$
Solving for $\tau$;
$$\tau=\dfrac{−30T}{2\ln(0.5)}\tag1 $$
Now we need to find the period which is given by
$$\omega=2\pi f=\sqrt{\dfrac{k}{m}}$$
where $f=1/T$, so
$$ \dfrac{2\pi}{T}=\sqrt{\dfrac{k}{m}}$$
Hence,
$$T=2\pi\sqrt{\dfrac{m}{k}} $$
Plug into (1) and then plug the known;
$$\tau=\dfrac{−30 \left(2\pi\sqrt{\dfrac{m}{k}}\right)}{2\ln(0.5)}=\dfrac{−30 \left(2\pi\sqrt{\dfrac{0.50}{15}}\right)}{2\ln(0.5)} $$
$$\tau =\color{red}{\bf 24.8}\;\rm s$$
