Answer
(a) $W_{net} = -2.15 \times 10^{-5} \mathrm{J}$
(b) $V = 2.83 \times 10^{3} \mathrm{V}$
(c) $|E| = 35.4 \times 10^3 \mathrm{V} / \mathrm{m}$
Work Step by Step
(a) The work done gains the particle kinetic energy
$$W=K=4.35 \times 10^{-5} \mathrm{J}$$
Also, the additional force applied by $6.5 \times 10^{-5} \,\text{J}$ but in the opposite direction. So, its value is negative. Hence, the net work done on the particle is
$$W_{net} = 4.35 \times 10^{-5} \mathrm{J} - 6.5 \times 10^{-5} \mathrm{J} = \boxed{-2.15 \times 10^{-5} \mathrm{J} }$$
The negative sign indicates that the particle will move in the direction of the additional force. (right)
(b) The potential difference is given by
$$V=\dfrac{W_{net}}{q}=\dfrac{-2.15 \times 10^{-5}}{7.6 \times 10^{-9}}=\boxed{-2.83 \times 10^{3} \mathrm{V}}$$
(c) The magnitude of the electric field is
$$|E|=\dfrac{|V|}{d}=\frac{|-2.83 \times 10^{3}|}{0.08}= \boxed{35.4 \times 10^3 \mathrm{V} / \mathrm{m}}$$
