Answer
a) $ 6.25 \times 10^8 \, \text{s}^{-1} $
b) $ \approx 0.17 \, \text{ns} $
Work Step by Step
$$\color{blue}{\bf [a]}$$
The decay rate $ r $ is calculated as the inverse of the lifetime $ \tau $:
$$
r = \frac{1}{\tau}\tag 1
$$
From the mentioned table, the lifetime of the $2p$ state of hydrogen is $ \tau = 1.6 \times 10^{-9} \, \text{s} $
Substitute into (1);
$$
r = \frac{1}{1.6 \times 10^{-9}} $$
$$
r= \color{red}{\bf 6.25 \times 10^8} \, \text{s}^{-1}
$$
$$\color{blue}{\bf [b]}$$
If 10% decay, then it means that 90% of the atoms remain excited, so:
$$
N_{\text{exc}} = 0.90 N_0
$$
Using the exponential decay equation:
$$
N_{\text{exc}} = N_0 e^{-t/\tau}
$$
Thus,
$$
0.90 \color{red}{\bf\not}N_0 = \color{red}{\bf\not}N_0 e^{-t/\tau}
$$
Taking the natural logarithm of both sides:
$$
\ln(0.90) = -\frac{t}{\tau}
$$
Solving for $ t $:
$$
t = -\tau \ln(0.90)
$$
Substitute the known;
$$
t = -(1.6 \times 10^{-9}) \ln(0.90)=\color{red}{\bf 1.69 \times 10^{-10}}\;\rm s
$$
