You have the hint: "you can take $k$ to be an estimate of the initial relative growth rate".
Since, $\text{growth rate} = \text{birth rate} - \text{death rate}$, we get that the growth rate is ranged from $15$ to $35$ million per year. Hence, the average growth rate is $\frac{15+35}{2} = 25$ million per year. Therefore, the initial relative growth rate $k = \frac{25}{5 300} = 0,0047169$.
I think it is what you need.
Since the population increases at a rate proportional to the number of people present at time t (read: the rate of change of $P$ is proportional to $P$.)
$$\frac{\mathrm{d}P}{\mathrm{d}t} = kP$$
Separating the variables and integrating yields
$$\int \frac{1}{P} \, \mathrm{d}P = \int k \, \mathrm{d}t$$
So that we get
$$\ln P = kt + c$$ or equivalently, letting the initial population be $P_0$ so that when $t=0$, $P= P_0$ to find the arbitrary constant and re-arranging yields
$$P = P_0e^{kt}$$
We are given, that the population doubles ($2P_0$) in five years ($t=5$) so:
$$2P_0 = P_0 e^{5k}$$
Solving for $k$ yields
$$e^{5k} = 2 \implies k = \frac{\ln 2}{5}$$
Our population growth equation is then
$$P = P_0 e^{\frac{\ln 2}{5}t}$$
So for the population to triple, we need to solve $3P_0 = P_0e^{kt}$ for $t$, where $k$ is what we found above. This yields
$$e^{kt} = 3 \implies kt = \ln 3 \implies t = \frac{5\ln 3}{\ln 2}$$
The same can be done to find the time it takes for the population to quadruple, that is solve $4P_0 = P_0e^{kt}$ for t.
For the second question, is we have that the population after three years ($t=3$) is $P = 10000$
then substituting this into our population growth equation gives
$$10000 = P_0e^{3k}$$
so that we solve for $P_0$ to get
$$P_0 = \frac{10000}{e^{3k}} = \frac{10000}{\exp{\left(\frac{3\ln 2}{5}\right)}}$$
Best Answer
Your model should have only two parameters, $\mu$ and $K$. The last term is for a harvest proportional to the population size, which does not play a role in this scenario.
Thus at $P=80\cdot 10^6$ you get $\mu P=0.7⋅10^6$ and $\frac{\mu P^2}K=0.9⋅10^6$ so that $\mu=\frac{7}{800}$ and $K=\frac{7⋅80}{9}⋅10^6\approx 62.2⋅10^6$.
Thus the current population is above carrying capacity, it will further shrink by having more deaths than births towards the carrying capacity of $62\,222\,222$