This is a cross-sectional view of candy bar ABC. A candy company wants to create a cylindrical container for candy bar ABC so that it is circumscribed about the candy bar. If segment AD = 4 cm, what is the smallest diameter of wrapper that will fit the candy bar?

To determine the smallest diameter of a cylindrical wrapper that can fit around the candy bar ABC, we first need to analyze the dimensions of the triangle based on the given information.

We have:
- Segment AD = 4 cm.
- Since the angles AED, ABF, and DFC are right angles, we can assume that AB and AC are both bisected.

From the given triangle ABC:
- Let AE = EB = x.
- Since AE = EB and points D and F are midpoints of segments AC and BC, respectively, we can conclude that:
- AD = DC = 4 cm, which implies that AC = AD + DC = 4 cm + 4 cm = 8 cm for the vertical side.

To make sense of triangle ABC:
- The rectangle AEF is inscribed in the triangle ABC. Therefore, point C must be somewhere that maintains a right triangle, making it symmetric about the midpoint.

Since AE = EB (and congruent lengths) were defined, we can calculate the widest distance across the triangle:
- Look into segment lengths vertically and horizontally.

In the right triangle framework:
- The maximum width can be calculated using the Pythagorean theorem. If AE = EB = x, then:
- \( EC = AE + EB = 2x \) vertically.
- The full horizontal length would depend on point D's position, where \( DE \) would also be defined symmetrically.

Now if \( AD = 4 \) cm, hence \( C \) spans out symmetrically bounded by it forming a right triangle along with AB. Thus:
- The diameter of the cylinder should essentially wrap around the highest vertex displacing symmetrically outwards.

To strictly find the minimum diameter necessary, we calculate the circular bound required which:
- Considering segment AD spans the way, with both heights and widths calculated from symmetry - effectively presenting an enclosed rectangle that perfectly matches the triangle angles, we'd need the maximal bi-dimensional distance across:

With \( AD = 4 cm \) giving us horizontal inclusions, the vertical max distance calculated reciprocally around each right angle vertex calculates \( sqrt(4^2 + 4^2) \), leading us to roughly estimate the required diameter via counting across intersections.

The final conclusion:
The radius = \( \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \).
Thus diameter would be \( 8\sqrt{2} \) concentrating the required cylindrical span around.

In essence to wrap on the strip, making a crucial analysis, proving diameter metrics run with calculated values, would suggest **8 cm** sufficing.

Thus, the correct answer for the smallest diameter of the wrapper that will fit the candy bar ABC is:
**8 cm.**