Answer
$R_{\text{eq}} = \dfrac{3R}{4}$
Work Step by Step
The circle has a total resistance $R/3$. and it is split into two resistors $R_{C1}$ and $R_{C2}$ which are equal, so each resistor equals $R/6$.
The combination of parallel $R_{C1}$ and $R_{C2}$ is
\begin{equation}
R_{\text{Ceq}} = \dfrac{R_{C1}R_{C2}}{R_{C1} + R_{C2} }
\end{equation}
Subsitutre the values for $R_{C1}$ and $R_{C2}$
\begin{align*}
R_{\text{Ceq}} = \dfrac{R_{C1}R_{C2}}{R_{C1} + R_{C2} } = \dfrac{(R/6)(R/6)}{R/6 + R/6 } = R/12
\end{align*}
$R_{1}$ and $R_{3}$ are in series with R_{\text{Ceq}}. So, the total equivalent resistance of the wire
\begin{align*}
R_{\text{eq}} = R_{1} + R_{\text{Ceq}} + R_{3} = R/3 + R/12 + R/3 = \boxed{\dfrac{3R}{4}}
\end{align*}
