A wire of resistance 15.0 is bent to form a regular hexagon ABCDEFA...

Total resistance of the wire is $15.0 \Omega$.

Since the wire is bent into a regular hexagon, it has $6$ equal sides. The resistance of each side is $R = \dfrac{15.0}{6} = 2.5 \Omega$.

The points $A$ and $B$ are adjacent vertices. The equivalent circuit consists of two parallel paths between $A$ and $B$:
1. The direct path along the side $AB$ with resistance $R_{1} = R = 2.5 \Omega$.
2. The path through the remaining five sides with resistance $R_{2} = 5R = 12.5 \Omega$.

The equivalent resistance $R_{eq}$ is given by the parallel combination of $R_{1}$ and $R_{2}$:

$R_{eq} = \dfrac{R_{1} R_{2}}{R_{1} + R_{2}} = \dfrac{2.5 \times 12.5}{2.5 + 12.5}$

$R_{eq} = \dfrac{31.25}{15.0} \approx 2.08 \Omega$