Calculate the cross-sectional area of a conductor and the total resistance of resistors in series.

Given resistivity ($\rho$) = $1.9 \times 10^{-8} \, \Omega \cdot \text{m}$.
Given length ($L$) = $20 \, \text{m}$.
Given resistance ($R$) = $0.4 \, \Omega$.
The formula relating resistance, resistivity, length, and cross-sectional area ($A$) is:

$$R = \rho \frac{L}{A}$$

$$A = (1.9 \times 10^{-8} \, \Omega \cdot \text{m}) \times \frac{20 \, \text{m}}{0.4 \, \Omega}$$ $$A = \frac{1.9 \times 10^{-8} \times 20}{0.4} \, \text{m}^2$$ $$A = \frac{38 \times 10^{-8}}{0.4} \, \text{m}^2$$ $$A = 95 \times 10^{-8} \, \text{m}^2$$ $$A = 9.5 \times 10^{-7} \, \text{m}^2$$

The cross-sectional area of the conductor is $\boxed{\text{9.5} \times 10^{-7} \, \text{m}^2}$.

** (Circuit with three resistors):**

Given resistors: $R_1 = 2 \, \Omega$, $R_2 = 4 \, \Omega$, $R_3 = 6 \, \Omega$.
For resistors in series, the total resistance ($R_{total}$) is the sum of individual resistances:

$$R_{total} = R_1 + R_2 + R_3$$ $$R_{total} = 2 \, \Omega + 4 \, \Omega + 6 \, \Omega$$ $$R_{total} = 12 \, \Omega$$

Given total voltage ($V$) = $12 \, \text{V}$.
According to Ohm's Law, $V = I R_{total}$, where $I$ is the total current.

$$I = \frac{V}{R_{total}}$$ $$I = \frac{12 \, \text{V}}{12 \, \Omega}$$ $$I = 1 \, \text{A}$$

The current flowing through the circuit is $\boxed{\text{1 A}}$.

** (Circuit with two resistors in parallel):**

Given resistors: $R_1 = 3 \, \Omega$, $R_2 = 6 \, \Omega$.
For resistors in parallel, the reciprocal of the equivalent resistance ($R_{eq}$) is the sum of the reciprocals of individual resistances:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$ $$\frac{1}{R_{eq}} = \frac{1}{3 \, \Omega} + \frac{1}{6 \, \Omega}$$ $$\frac{1}{R_{eq}} = \frac{2}{6 \, \Omega} + \frac{1}{6 \, \Omega}$$ $$\frac{1}{R_{eq}} = \frac{3}{6 \, \Omega}$$ $$\frac{1}{R_{eq}} = \frac{1}{2 \, \Omega}$$ $$R_{eq} = 2 \, \Omega$$

Given total current ($I_{total}$) = $4 \, \text{A}$.
According to Ohm's Law, $V = I_{total} R_{eq}$, where $V$ is the total voltage.

$$V = (4 \, \text{A}) \times (2 \, \Omega)$$ $$V = 8 \, \text{V}$$

The voltage applied to the circuit is $\boxed{\text{8 V}}$.