Two point charges, A and B, each with a charge of +3 x 10⁻⁹ C, are stationary on a horizontal surface. Point P is r metres from charge A and 2r metres from charge B, as shown in the diagram below. Calculate the value of r.

The electric field $E$ due to a point charge $Q$ at a distance $d$ is given by:

$$E = \frac{kQ}{d^2}$$

where $k$ is Coulomb's constant, $k = 9 \times 10^9 \text{ N} \cdot \text{m}^2 \cdot \text{C}^{-2}$.
Given:
$Q_A = +3 \times 10^{-9} \text{ C}$
$Q_B = +3 \times 10^{-9} \text{ C}$
Distance from A to P is $d_A = r$.
Distance from B to P is $d_B = 2r$.
Net electric field at P is $E_{net} = 27 \text{ N} \cdot \text{C}^{-1}$.

Since both charges are positive, the electric field $E_A$ due to charge A at point P will point away from A (to the right). The electric field $E_B$ due to charge B at point P will point away from B (to the left).

$$E_A = \frac{kQ_A}{r^2} = \frac{(9 \times 10^9 \text{ N} \cdot \text{m}^2 \cdot \text{C}^{-2})(3 \times 10^{-9} \text{ C})}{r^2} = \frac{27}{r^2} \text{ N} \cdot \text{C}^{-1}$$ $$E_B = \frac{kQ_B}{(2r)^2} = \frac{(9 \times 10^9 \text{ N} \cdot \text{m}^2 \cdot \text{C}^{-2})(3 \times 10^{-9} \text{ C})}{4r^2} = \frac{27}{4r^2} \text{ N} \cdot \text{C}^{-1}$$

Since $E_A$ and $E_B$ are in opposite directions, the net electric field is their difference. As $r < 2r$, $E_A$ will be greater than $E_B$.

$$E_{net} = E_A - E_B$$ $$27 \text{ N} \cdot \text{C}^{-1} = \frac{27}{r^2} - \frac{27}{4r^2}$$ $$27 = 27 \left( \frac{1}{r^2} - \frac{1}{4r^2} \right)$$ $$1 = \frac{1}{r^2} \left( 1 - \frac{1}{4} \right)$$ $$1 = \frac{1}{r^2} \left( \frac{3}{4} \right)$$ $$r^2 = \frac{3}{4}$$

$$r = \sqrt{\frac{3}{4}}$$ $$r = \frac{\sqrt{3}}{2} \text{ m}$$ $$r \approx 0.866 \text{ m}$$

The value of $r$ is $\boxed{0.866 \text{ m}}$.

The electrostatic force $F$ experienced by a charge $q$ in an electric field $E$ is given by:

$$F = qE$$

Charge of an electron $q_e = -1.6 \times 10^{-19} \text{ C}$.
Magnitude of the net electric field at P is $E_{net} = 27 \text{ N} \cdot \text{C}^{-1}$.

$$F = |q_e| \times E_{net}$$ $$F = (1.6 \times 10^{-19} \text{ C}) \times (27 \text{ N} \cdot \text{C}^{-1})$$ $$F = 43.2 \times 10^{-19} \text{ N}$$ $$F = 4.32 \times 10^{-18} \text{ N}$$

The magnitude of the net electrostatic force is $\boxed{4.32 \times 10^{-18} \text{ N}}$.