Explanation:
To prove that option 'B' is the correct answer, we need to show that if G/Z is cyclic, then G is abelian.

Definition:
Let's start by defining some terms:
- Z(G) is the center of the group G, which is the set of elements that commute with every element of G. In other words, Z(G) = {z ∈ G | zg = gz for all g ∈ G}.
- G/Z is the quotient group of G by Z, which is the set of cosets of Z in G. In other words, G/Z = {gZ | g ∈ G}.

Proof:
We will prove that if G/Z is cyclic, then G is abelian.

Step 1: Assume G/Z is cyclic.
Since G/Z is cyclic, there exists an element gZ ∈ G/Z such that every element of G/Z can be expressed as a power of gZ. Let's denote this element as gZ = (gZ)^k for some integer k.

Step 2: Express elements of G/Z in terms of cosets.
For any element x ∈ G, we can express the coset xZ as xZ = (gZ)^m for some integer m.

Step 3: Express elements of G in terms of cosets.
Using the coset notation, we can express any element x ∈ G as x = gz for some g ∈ G and z ∈ Z.

Step 4: Simplify the expression.
Now, let's substitute the expression for x in terms of cosets into the expression for cosets in terms of gZ:
gz = (gZ)^m

Step 5: Use the definition of Z(G).
Since z ∈ Z(G), we know that gz = zg for all z ∈ Z(G). Therefore, we can rewrite the expression as:
zg = (gZ)^m

Step 6: Simplify the expression further.
Using the properties of cosets, we can rewrite the expression as:
gZ = (gZ)^m

Step 7: Deduce that m = 1.
Since every element of G/Z can be expressed as a power of gZ, we can conclude that m = 1.

Step 8: Conclude that G is abelian.
If m = 1, then x = gz = zg for all x ∈ G and z ∈ Z(G). Therefore, every element of G commutes with every element of G, which means G is abelian.

Conclusion:
Based on the proof above, if G/Z is cyclic, then G is abelian. Therefore, option 'B' is the correct answer.