Answer
${\bf 5.72}\, \text{ns}$
Work Step by Step
The decay of excited atoms is given by
$$
N_{\text{exc}}(t) = N_0 e^{-t/\tau}
$$
We are given that 1% of the atoms decay in 0.20 ns. In other words, after 0.20 ns, 99% of the atoms are still in the excited state.
So,
$$
0.99 \color{red}{\bf\not} N_0 = \color{red}{\bf\not} N_0 e^{-0.20 \, \text{ns} / \tau}
$$
Take the natural logarithm of both sides:
$$
\ln(0.99) = - \frac{0.20 \, \text{ns}}{\tau}
$$
$$
\tau = \frac{0.20 \, \text{ns}}{-\ln(0.99)}=\bf 19.9\;\rm ns
$$
Thus, the characteristic lifetime $ \tau $ of the excited state is approximately 19.90 ns.
Next, we need to find the time it will take for 25% of the sample to decay, meaning that 75% of the atoms will remain in the excited state. Using the same exponential decay equation:
$$
0.75 \color{red}{\bf\not} N_0 = \color{red}{\bf\not} N_0 e^{-t / \tau}
$$
Take the natural logarithm of both sides:
$$
\ln(0.75) = - \frac{t}{\tau}
$$
Solve for $ t $:
$$
t = -\tau \ln(0.75)=-19.90 \ln(0.75)
$$
Substitute $ \tau = 19.90 \, \text{ns} $ and $ \ln(0.75) \approx -0.28768 $:
$$
t \approx \color{red}{\bf 5.72}\, \text{ns}
$$
