**Solution:**

Let's assume the principal sum of money initially invested is P.

**Step 1:** Calculating the compound interest for 3 years.
The formula for compound interest is given by:
\[A = P\left(1 + \frac{r}{100}\right)^n\]
where A is the total amount after n years, P is the principal, and r is the rate of interest.

Given that the total compound interest earned at the end of 3 years is ₹2,170, we can write the equation as:
\[A = P\left(1 + \frac{r}{100}\right)^3\]
\[A - P = P\left(1 + \frac{r}{100}\right)^3 - P\]
\[2170 = P\left(\left(1 + \frac{r}{100}\right)^3 - 1\right)\] ---(Equation 1)

**Step 2:** Calculating the compound interest for 2 years.
Using the same formula, we can calculate the total amount after 2 years:
\[A = P\left(1 + \frac{r}{100}\right)^2\]
\[A - P = P\left(1 + \frac{r}{100}\right)^2 - P\]
\[1360 = P\left(\left(1 + \frac{r}{100}\right)^2 - 1\right)\] ---(Equation 2)

**Step 3:** Solving the equations.
Now, let's solve equations (1) and (2) to find the values of P and r.

Dividing equation (1) by equation (2):
\[\frac{2170}{1360} = \frac{P\left(\left(1 + \frac{r}{100}\right)^3 - 1\right)}{P\left(\left(1 + \frac{r}{100}\right)^2 - 1\right)}\]
\[\frac{2170}{1360} = \frac{\left(1 + \frac{r}{100}\right)^3 - 1}{\left(1 + \frac{r}{100}\right)^2 - 1}\]

Simplifying the equation, we get:
\[\frac{2170}{1360} = \frac{\left(1 + \frac{r}{100}\right)^3 - 1}{\left(1 + \frac{r}{100}\right)^2 - 1}\]
\[\frac{2170}{1360} = \frac{\left(1 + \frac{r}{100}\right)^3 - 1}{\left(1 + \frac{r}{100}\right) - 1}\]
\[\frac{2170}{1360} = \frac{\left(1 + \frac{r}{100}\right)^3 - 1}{\frac{r}{100}}\]
\[\frac{2170}{1360} = \frac{1 + 3\left(\frac{r}{100}\right) + 3\left(\frac{r}{100}\right)^2 + \left(\frac{r}{100}\right)^3 - 1}{\frac{r}{100}}\]
\[\frac{2170}{136