Step 1: Understand the idea.
Simple Interest stays the same each year, but Compound Interest grows because you earn interest on past interest. So Compound Interest is always a little more than Simple Interest, and we want that small difference.
Step 2: Use the 3-year difference formula.
For $3$ years, the gap between Compound Interest and Simple Interest is \[ \text{CI}-\text{SI}=P\left(\frac{r}{100}\right)^2\left(3+\frac{r}{100}\right), \] where $P$ is the principal and $r$ is the yearly rate.
Step 3: Put in the given values.
Here $r=10$ and the difference is $155$. So \[ 155=P\left(\frac{10}{100}\right)^2\left(3+\frac{10}{100}\right). \]
Step 4: Simplify the bracket terms.
Note $\frac{10}{100}=\frac{1}{10}$, so $\left(\frac{1}{10}\right)^2=\frac{1}{100}$, and $3+\frac{1}{10}=\frac{31}{10}$. Then \[ 155=P\times\frac{1}{100}\times\frac{31}{10}=P\times\frac{31}{1000}. \]
Step 5: Solve for the principal $P$.
Multiply both sides by $\frac{1000}{31}$: \[ P=155\times\frac{1000}{31}. \] Since $155=31\times5$, this gives \[ P=5\times1000=5000. \]
Step 6: State the answer.
The principal sum is rupees $5000$. \[ \boxed{\text{Rs }5000} \]