Let N be the number of bacteria at time t (in hours). Since the rate of growth is proportional to the number present, we have the differential equation:

$$\frac{dN}{dt} = kN$$

where k is the constant of proportionality.

Separating variables:

$$\frac{dN}{N} = k\, dt$$

Integrating both sides:

$$\int \frac{dN}{N} = \int k\, dt \Rightarrow \ln N = kt + C$$

where $C$ is the constant of integration. Exponentiating:

$$N = e^{kt + C} = e^C \, e^{kt} = N_0 \, e^{kt}$$

where $N_0 = e^C$ is the initial count.

Given: at $t = 0$, $N = 1{,}00{,}000$, so $N_0 = 1{,}00{,}000$. Hence,

$$N = 1{,}00{,}000 \, e^{kt}$$

Also given: the count increases by $10\%$ in 2 hours, i.e., at $t = 2$, $N = 1{,}10{,}000$. Substitution:

$$1{,}10{,}000 = 1{,}00{,}000 \, e^{2k} \Rightarrow e^{2k} = \frac{11}{10}$$

Taking natural logarithm:

$$2k = \ln\!\left(\frac{11}{10}\right) \Rightarrow k = \frac{1}{2} \ln\!\left(\frac{11}{10}\right)$$

We need the time T when $N = 2{,}00{,}000$. So:

$$2{,}00{,}000 = 1{,}00{,}000 \, e^{kT} \Rightarrow e^{kT} = 2 \Rightarrow kT = \ln 2$$ $$T = \frac{\ln 2}{k} = \frac{\ln 2}{\frac{1}{2} \ln\!\left(\frac{11}{10}\right)} = \frac{2 \ln 2}{\ln\!\left(\frac{11}{10}\right)} \text{ hours}$$

∴ The bacteria count will reach 2,00,000 in $T = \dfrac{2 \ln 2}{\ln(11/10)}$ hours.