Answer
$$ N'(t)=\frac{\ln2}{3}N(t).$$
$$ N(t)=P_0e^{kt}=e^{t\frac{\ln 2}{3}}.$$
$$ N(60)= e^{20\ln 2}.$$
Work Step by Step
Since the doubling time is given by $\frac{\ln 2}{k}$, then we have
$$3=\frac{\ln 2}{k}\Longrightarrow k=\frac{\ln2}{3}.$$
The differential equation is given by
$$ N'(t)=kN(t)\Longrightarrow N'(t)=\frac{\ln2}{3}N(t).$$
When $ t=0$, we have $ P_0=N(0)=1$, then
$$ N(t)=P_0e^{kt}=e^{t\frac{\ln 2}{3}}.$$
At $ t=60$, we have
$$ N(60)= e^{60\frac{\ln 2}{3}}=e^{60\frac{\ln 2}{3}}=e^{20\ln 2}.$$
