1. Form the differential equation

Since the rate of growth is proportional to the number of bacteria present, $\frac{dN}{dt}=rN$ for some constant $r>0$.

Its solution is $N(t)=N_0 e^{rt}$ Given $N(0)=1000$, we get $N(t)=1000e^{rt}$

2. Use the 20% increase in 2 hours

After 2 hours, the population becomes $120\%$ of 1000: $N(2)=1200$ So, $1000e^{2r}=1200$ $e^{2r}=\frac{6}{5}$ Taking natural log, $2r=\ln\left(\frac{6}{5}\right)$ $r=\frac{1}{2}\ln\left(\frac{6}{5}\right)$

3. Use the condition when population becomes 2000

We are told that the population is 2000 after $\frac{k}{\ln(6/5)}$ hours.

So, $N\left(\frac{k}{\ln(6/5)}\right)=2000$

Using $N(t)=1000e^{rt}$, $1000e^{r\cdot \frac{k}{\ln(6/5)}}=2000$ $e^{r\cdot \frac{k}{\ln(6/5)}}=2$ Taking natural log, $r\cdot \frac{k}{\ln(6/5)}=\ln 2$

Now substitute $r=\frac{1}{2}\ln\left(\frac{6}{5}\right)$ Then, $\frac{1}{2}\ln\left(\frac{6}{5}\right)\cdot \frac{k}{\ln(6/5)}=\ln 2$ $\frac{k}{2}=\ln 2$ $k=2\ln 2$

4. Compute the required quantity

We need $\left(\frac{k}{\ln 2}\right)^2$ Substitute $k=2\ln 2$: $\left(\frac{2\ln 2}{\ln 2}\right)^2=2^2=4$

5. Final answer

$\boxed{4}$ So the correct option is D.