Coming up with any sort of answer here will require a large number of assumptions. The most important factor is how the growth rate changes with population. The one solid value you've given is that the average lifespan is 100 years. I will make the simplifying assumption that this is from all causes, so includes the impact of unnatural deaths. That means that $1\%$ of the population dies every year. At $10^{10} = 10$ billion people, the growth rate is $0\%$, so the birth rate must be $1\%$ as well.
On the other end of the spectrum, you mention "5-10 range", which I assume means 5 to 10 children per couple for small colonies. 10 children amounts to a $5\%$ birthrate per year (5 kids per adult over a 100 year lifespan), and a growth rate of $5\% - 1\% = 4\%$. Let's assume that is the growth rate at $0$ population (note that it is an average birthrate, so the childless are accounted in that average).
To make the math simple, let's just assume a linear decrease in growthrate from $4\%$ at $0$ population to $0\%$ at $10$ billion population. So the yearly growth rate at population $P$ (measured in billons) is $$0.04 - \frac {0.04}{10}P$$ This gives a differential equation $$\frac {dP}{dt} = 0.04P - 0.004P^2$$
The solution to this equation is $$P = 10\frac {Re^{0.04t}}{1+Re^{0.04t}}$$ where $R = \frac {P_0}{10 - P_0}$ and $P_0$ is the initial population (in billions, at $t = 0$).
Now the less population a planet has, the faster it produces more. So planets should keep their own population as low as possible, to take advantage of the greater rate. Thus any planet with a population of $5$ billion should ship additional people away to its colony at a rate matching its growth ($2\%$, or $0.1$ billion a year), thus maintaining its population at the minimum necessary to have a colony. This adds an additional factor to the growthrate of the colony:
$$\frac {dP}{dt} = 0.1 + 0.04P - 0.004P^2$$
If we set $P = 0$ at $t = 0$ (since the parent colony is shipping $100$ million people a year, the requisite $10,000$ will be reached in the first day), the solution to this is $$P = 5 +5\sqrt{2}\frac{\sqrt 2 - 1 - (\sqrt 2 + 1)e^{-rt}}{\sqrt 2 - 1 + (\sqrt 2 + 1)e^{-rt}}$$ where $r = \frac{\sqrt 2}{25}$. Setting $P = 5$ and solving for $t$ gives that colony will need about $31$ years, $2$ months to reach $5$ billion in population.
At this point the total population has doubled, and now we have two planets at $5$ billion who can start two new colonies, repeating the process. I.e., the total population will double every $31$ years, $2$ months. To reach one trillion from $5$ billion requires $7.64$ doublings, so one should reach $1$ trillion by $238$ years (straight multiplying like that is not exactly accurate, but gives a quick estimation and is a relatively minor error compared to all the assumptions I've made).
Because Earth starts with $7$ billion instead of $5$, the first colony gets a head start, beginning at $2$ billion instead of $0$. That will several years off the initial $31$. But again, there is enough other error here to let that slide, so I will leave the estimate at $238$ years.
Best Answer
So the growth function for the population of China is $C(t)=1.34(1.004)^t$, and for India $I(t)=1.19(1.0137)^t$. So, we need to solve the inequality
$$\begin{align} 1.19(1.0137)^t & > 1.34(1.004)^t\\ \left(\frac{1.19}{1.34}\right)\left(\frac{1.0137}{1.004}\right)^t &>1\\ \left(\frac{1.0137}{1.004}\right)^t & > \left(\frac{1.34}{1.19}\right)\\ t\log\left(\frac{1.0137}{1.004}\right)& > \log\left(\frac{1.34}{1.19}\right)\\ t & > {\log(1.34/1.19)\over\log(1.0137/1.004)}\\ t& > 12.35 \end{align}$$