Answer
(a) Point $b$
(b) $800 \mathrm{~V} / \mathrm{~m}$
(c) $W = 48 \times 10^{-6} \mathrm{J}$
Work Step by Step
(a) As shown by the values of $a$ and $b$, the electric field points from the $b$ to $a$ where $b$ is in the +x-axis far than $a$. The electric field points from the higher energy to the lower energy, therefore, the higher energy at $$\boxed{\text{point}\,\, b}$$.
(b) The electric field is related to the distance between the two charges by
$$E=\dfrac{V_{a b}}{d}=\dfrac{240 \mathrm{V}}{0.9 \,\text{m} -0.6 \,\text{m}}=\boxed{800 \mathrm{~V} / \mathrm{~m}}$$
(c) The work done is related to the electric field by
$$W= Fd = E q d=(800 \mathrm{V} / \mathrm{m})\left(-0.2 \times 10^{-6}\right)(0.3 \mathrm{m})=-48 \times 10^{-6} \mathrm{J}$$
