Answer
(a) $V_a$ = 0
(b) $V_b = - 171\times 10^{3} \,\text{V}$
(c) $W = -0.855\mathrm{J} $
Work Step by Step
(a) The potential at point $a$ will be the potential between the two charges
\begin{align}
V_{a} &= \dfrac{1}{4\pi \epsilon_o} \sum \frac{q}{r_{i}}\\
&= \dfrac{1}{4\pi \epsilon_o} \left(\frac{q_{1}}{r}+\frac{q_{2}}{r}\right)\\
& = \dfrac{1}{4\pi \epsilon_o} \left(\frac{+2.00 \mu \mathrm{C}}{r}+\frac{-2.00 \mu \mathrm{C}}{r}\right) \\
& = 0
\end{align}
(b) From Physagros theorem, we could get the distance between $q_1$ and point $b$ by
$$r_{1}=\sqrt{(.03)^{2}+(.03)^{2}}=0.042 \mathrm{m}$$
Now we could get the potential at point $b$ by
\begin{align}
V_{b} &= \dfrac{1}{4\pi \epsilon_o} \sum \frac{q}{r_{i}}\\
&= \dfrac{1}{4\pi \epsilon_o} \left(\frac{q_{1}}{r_1}+\frac{q_{2}}{r_2}\right)\\
& = \dfrac{1}{4\pi \epsilon_o} \left(\frac{+2.00 \mu \mathrm{C}}{0.042 \mathrm{m}}+\frac{-2.00 \mu \mathrm{C}}{0.03 \mathrm{m}}\right) \\
& =\boxed{ - 171\times 10^{3} \,\text{V}}
\end{align}
(c) The work done $W$ is given by
\begin{aligned}
W &=q_{3} \Delta V\\
& = q_3 \left(V_{a}-V_{b}\right) \\
&=\left(-5 \times 10^{-6}\right)\left(0 - [-171 \times 10^{3}]\right)\\
&=\boxed{-0.855 \mathrm{J} }
\end{aligned}
