Answer
(a) $W =$ 3.4 J.
(b) No. It is not.
(c) $v$ = 67 m/s
Work Step by Step
(a) The distance required is the radius of the ring and equal $r = d/2 $ = 0.08 m /2 = 0.04 m
The work done between the two charges is given by
\begin{aligned}
W &=\frac{1}{4 \pi \epsilon_{0}} \frac{q_1 q_2}{R} \\ &=\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left(3.00 \times 10^{-6} \mathrm{C}\right)\left(5.00 \times 10^{-6} \mathrm{C}\right)}{0.04 \mathrm{m}} \\ &=\boxed{ 3.4 \mathrm{J}}
\end{aligned}
(b) The charges are distributed uniformly, so it is not necessary to take a path along the axis of the ring.
(c) The initial potential energy converts to a kinetic energy
\begin{gather}
U =\frac{1}{2} m v^{2} \\
v =\sqrt{\frac{2 U}{m}} \\
v =\sqrt{\frac{2 \times 3.4 \mathrm{J}}{0.0015 \mathrm{kg}}} \\
v =\boxed{67 \mathrm{m} / \mathrm{s} }
\end{gather}
