A charge of +6.0 µC is at the origin. A charge of -4.0 µC is at (0.30 m, 0)…

Step 1. From the $+6.0\ \mu\text{C}$ charge. The distance is $0.40$ m. $F_1 = \dfrac{kq_Aq}{r^2}$. The top is $9.0 \times 10^9 \times 6.0 \times 10^{-6} \times 1.5 \times 10^{-6}\ \text{N m}^2 = 0.081\ \text{N m}^2$ and the bottom is $(0.40\ \text{m})^2 = 0.16\ \text{m}^2$. So $F_1 = 0.506\ \text{N}$. It is a push, along $+y$.

Step 2. From the $-4.0\ \mu\text{C}$ charge. The distance is $\sqrt{(0.30\ \text{m})^2 + (0.40\ \text{m})^2} = 0.50$ m. $F_2 = \dfrac{kq_Bq}{r^2}$. The top is $9.0 \times 10^9 \times 4.0 \times 10^{-6} \times 1.5 \times 10^{-6}\ \text{N m}^2 = 0.054\ \text{N m}^2$ and the bottom is $(0.50\ \text{m})^2 = 0.25\ \text{m}^2$. So $F_2 = 0.216\ \text{N}$. It is a pull, towards the $-4.0\ \mu\text{C}$ charge, down and to the right.

Step 3. The components of $F_2$. The direction has $\cos\alpha = 0.30/0.50 = 0.60$ and $\sin\alpha = 0.40/0.50 = 0.80$. So $F_{2x} = +0.216\ \text{N} \times 0.60 = +0.130\ \text{N}$ and $F_{2y} = -0.216\ \text{N} \times 0.80 = -0.173\ \text{N}$.

Step 4. The sums. $F_x = 0.130\ \text{N}$ and $F_y = 0.506\ \text{N} - 0.173\ \text{N} = 0.333\ \text{N}$.

Step 5. The net force. $F = \sqrt{(0.130\ \text{N})^2 + (0.333\ \text{N})^2} = 0.358\ \text{N}$, and $\tan\theta = \dfrac{0.333\ \text{N}}{0.130\ \text{N}} = 2.57$, so $\theta = 68.8^\circ$.