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(B) The electric potential $V$ due to a charged arc of radius $R$ and charge $Q$ at its center is given by $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}$

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$\frac{Q}{{2\pi { \in _0}R}}$

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(B) The electric potential $V$ due to a charged arc of radius $R$ and charge $Q$ at its center is given by $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}$.Since the potential is a scalar quantity,the net potential at the center is the algebraic sum of the potentials due to the three arcs.$V_{net} = V_1 + V_2 + V_3$$V_{net} = \frac{1}{4\pi \epsilon_0 R} (Q_1 + Q_2 + Q_3)$Given charges are $Q_1 = +Q$,$Q_2 = +3Q$,and $Q_3 = -2Q$.$V_{net} = \frac{1}{4\pi \epsilon_0 R} (Q + 3Q - 2Q)$$V_{net} = \frac{1}{4\pi \epsilon_0 R} (2Q)$$V_{net} = \frac{2Q}{4\pi \epsilon_0 R} = \frac{Q}{2\pi \epsilon_0 R}$.

The figure shows three circular arcs,each of radius $R$ and total charge as indicated. The net electric potential at the centre of curvature is

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