1. **State the problem:** We are given a geometric progression (GP) with the first term $a=3$ and common ratio $r=3$. We need to find which term of this series is equal to 243. 2. **Formula for the $n$th term of a GP:** $$a_n = a \times r^{n-1}$$ where $a_n$ is the $n$th term, $a$ is the first term, $r$ is the common ratio, and $n$ is the term number. 3. **Apply the formula:** We want to find $n$ such that $$a_n = 243$$ Substitute $a=3$ and $r=3$: $$3 \times 3^{n-1} = 243$$ 4. **Simplify the equation:** $$3^{1} \times 3^{n-1} = 3^{5}$$ Using the property of exponents $3^{1} \times 3^{n-1} = 3^{n}$, so $$3^{n} = 3^{5}$$ 5. **Equate the exponents:** Since the bases are equal and non-zero, the exponents must be equal: $$n = 5$$ 6. **Conclusion:** The term 243 is the 5th term of the geometric progression. **Final answer:** $n=5$