A wire possessing a resistance of 15.0 is bent into a regular...

Total resistance of the wire, $R = 15.0 \Omega$

Since the wire is bent into a regular hexagon, it has $6$ equal sides.

Resistance of each side $= \dfrac{15.0}{6} = 2.5 \Omega$

The equivalent resistance is to be found between points $A$ and $C$. The hexagon can be divided into two parallel paths between $A$ and $C$:

Path 1: $A \rightarrow B \rightarrow C$ ($2$ sides)

Resistance of Path 1, $R_1 = 2 \times 2.5 = 5.0 \Omega$

Path 2: $A \rightarrow F \rightarrow E \rightarrow D \rightarrow C$ ($4$ sides)

Resistance of Path 2, $R_2 = 4 \times 2.5 = 10.0 \Omega$

Since $R_1$ and $R_2$ are in parallel, the equivalent resistance $R_{eq}$ is:

$R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2} = \dfrac{5.0 \times 10.0}{5.0 + 10.0} = \dfrac{50.0}{15.0} = \dfrac{10}{3} \Omega$

$R_{eq} \approx 3.33 \Omega$