Two large parallel plates are 5.0 cm apart, with 600 V across them…

Step 1. The field. $E = \dfrac{V}{d} = \dfrac{600\ \text{V}}{0.050\ \text{m}} = 12000\ \text{V/m}$.

Step 2. The distance along the field. $8.0\ \text{cm} \times \cos 60^\circ = 4.0\ \text{cm} = 0.040\ \text{m}$. The path runs against the field, so the potential rises.

Step 3. The potential difference. $V_Q - V_P = E \times 0.040\ \text{m} = 12000\ \text{V/m} \times 0.040\ \text{m} = 480\ \text{V}$.

Step 4. The work. $W = q\,\Delta V = 3.0 \times 10^{-9}\ \text{C} \times 480\ \text{V} = 1.44 \times 10^{-6}\ \text{J}$, that is $1.44\ \mu\text{J}$.