Concept:
If a code consists of \(n\) bits, the total number of distinct combinations possible is:
\[
2^n
\]
To represent a given number of elements, we must choose the smallest value of \(n\) such that:
\[
2^n \geq \text{Number of Elements}
\]
Step 1: Identify the number of elements.
The question states that there are:
\[
8
\]
distinct elements.
Step 2: Determine the number of bits required.
Check powers of 2:
\[
2^1=2
\]
\[
2^2=4
\]
\[
2^3=8
\]
\[
2^4=16
\]
We observe that:
\[
2^3=8
\]
Therefore exactly three bits can represent eight distinct elements.
Step 3: Verify using binary combinations.
Three bits generate:
\[
000,\;001,\;010,\;011,\;100,\;101,\;110,\;111
\]
Total combinations:
\[
8
\]
Hence a three-bit code is sufficient.
\[
\boxed{\text{Three-bit Code}}
\]
Therefore, option (A) is correct.