Figure shows a graph of $\ln \left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)$. From the graph, the half life (in seconds) of the radioactive sample is:

 

$$\begin{gathered} N=N_0{e}^{-\lambda t} \\ \Rightarrow \frac{dN}{dt}=-\lambda N_0{e}^{-\lambda t} \\ \therefore \left|\frac{dN}{dt}\right|=\lambda N_0{e}^{-\lambda t} \end{gathered}$$

Taking log on both sides, we get:

$\ln \left|\frac{dN}{dt}\right|=\ln \left(\lambda N_0\right)-\lambda t$

Slope of the graph is $-\lambda$, which is 

$$\begin{gathered} -\lambda =\frac{8-4}{2-6} \\ \Rightarrow \lambda =1 \end{gathered}$$

Now, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln \left(2\right)}{\lambda }=\frac{\ln \left(2\right)}{1}=\ln \left(2\right)$