Three point charges sit at the corners of a right-angled triangle…

Step 1. The pair AB. The product of the charges is $q_Aq_B = (2.0 \times 10^{-6}\ \text{C})(-3.0 \times 10^{-6}\ \text{C}) = -6.0 \times 10^{-12}\ \text{C}^2$. Then $U = k\dfrac{q_Aq_B}{r} = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2 \times \dfrac{-6.0 \times 10^{-12}\ \text{C}^2}{0.30\ \text{m}} = -0.18\ \text{J}$.

Step 2. The pair BC. $q_Bq_C = -12 \times 10^{-12}\ \text{C}^2$. $U = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2 \times \dfrac{-12 \times 10^{-12}\ \text{C}^2}{0.40\ \text{m}} = -0.27\ \text{J}$.

Step 3. The pair AC. $q_Aq_C = 8.0 \times 10^{-12}\ \text{C}^2$. $U = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2 \times \dfrac{8.0 \times 10^{-12}\ \text{C}^2}{0.50\ \text{m}} = +0.144\ \text{J}$.

Step 4. The group. $U = -0.18\ \text{J} - 0.27\ \text{J} + 0.144\ \text{J} = -0.306\ \text{J}$.

Step 5. Pulling them apart. The energy at infinity is zero, so the work needed is $0 - (-0.306\ \text{J}) = 0.306\ \text{J}$.