Answer
a) $\ln(2)\;\tau$
b) ${\bf 11.8} \, \text{ns}$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the half-life $ t_{1/2} $ is the time required for half of the initially excited atoms to decay.
$$
N_{\text{exc}}(t) = N_0 e^{-t/\tau}\tag 1
$$
This means that after the half-life $ t = t_{1/2} $, the number of excited atoms is half of the initial number:
$$
N_{\text{exc}}(t_{1/2}) = \frac{1}{2} N_0
$$
Plug into (1);
$$
\frac{1}{2} \color{red}{\bf\not}N_0= \color{red}{\bf\not}N_0 e^{-t/\tau}\tag 1
$$
Take the natural logarithm on both sides:
$$
\ln\left(\frac{1}{2}\right) = -\frac{t_{1/2}}{\tau}
$$
Since $ \ln\left(\frac{1}{2}\right) = -\ln(2) $, we have:
$$
-\ln(2) = -\frac{t_{1/2}}{\tau}
$$
Thus, the half-life is related to the lifetime $ \tau $ by:
$$
\boxed{t_{1/2} = \tau \ln(2) }
$$
$$\color{blue}{\bf [b]}$$
From Table 41.3, the lifetime $ \tau $ for the $ 3p $ state of sodium is $ \tau = 17 \, \text{ns} $.
Plug that into the boxed formula above.
$$
t_{1/2} = \ln(2) \times 17 \approx \color{red}{\bf 11.8} \, \text{ns}
$$
