Consider the forces on one sphere: tension ($T$), weight ($mg$), and electrostatic force ($F_e$). The sphere is in equilibrium. The electrostatic force is repulsive:

$$F_e = k \frac{q^2}{d^2} = (9.0 \times 10^9) \frac{(3.0 \times 10^{-7})^2}{(0.4)^2} = 5.0625 \times 10^{-3} N$$

Let $\theta$ be the angle the thread makes with the vertical. From geometry:

$$\sin\theta = \frac{d/2}{L} = \frac{0.4/2}{1.0} = 0.2$$ $$\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - 0.2^2} = \sqrt{0.96} \approx 0.9798$$

The horizontal and vertical components of the forces must balance:

1. Horizontal: $T\sin\theta = F_e$
2. Vertical: $T\cos\theta = mg$

From equation (1), we can find the tension $T$:

$$T = \frac{F_e}{\sin\theta} = \frac{5.0625 \times 10^{-3} N}{0.2} = 0.0253 N$$

From equation (2), we can find the mass $m$:

$$m = \frac{T\cos\theta}{g} = \frac{(0.0253 N)(\sqrt{0.96})}{9.8 m/s^2} \approx 2.53 \times 10^{-3} kg$$