Given that (3x+2) is a factor of P(x) = 18x³ + bx² − x − 2, we set 3x+2=0 to find the root.

The Factor Theorem states that if $(ax+b)$ is a factor of a polynomial $P(x)$, then $P\left(-\frac{b}{a}\right) = 0$.
Given that $(3x+2)$ is a factor of $P(x) = 18x^3 + bx^2 - x - 2$, we set $3x+2=0$ to find the root.

$$3x = -2$$ $$x = -\frac{2}{3}$$

According to the Factor Theorem, $P\left(-\frac{2}{3}\right)$ must be equal to 0.

Calculate the powers:

$$\left(-\frac{2}{3}\right)^3 = -\frac{2^3}{3^3} = -\frac{8}{27}$$ $$\left(-\frac{2}{3}\right)^2 = \frac{2^2}{3^2} = \frac{4}{9}$$

Substitute these values back into the equation:

$$18\left(-\frac{8}{27}\right) + b\left(\frac{4}{9}\right) - \left(-\frac{2}{3}\right) - 2 = 0$$ $$-\frac{18 \cdot 8}{27} + \frac{4b}{9} + \frac{2}{3} - 2 = 0$$

Simplify the first term: $\frac{18}{27} = \frac{2 \cdot 9}{3 \cdot 9} = \frac{2}{3}$.
So, $-\frac{2 \cdot 8}{3} = -\frac{16}{3}$.
The equation becomes:

$$-\frac{16}{3} + \frac{4b}{9} + \frac{2}{3} - 2 = 0$$

Combine the constant terms:

$$\left(-\frac{16}{3} + \frac{2}{3}\right) + \frac{4b}{9} - 2 = 0$$ $$-\frac{14}{3} + \frac{4b}{9} - 2 = 0$$

To eliminate the denominators, multiply the entire equation by the least common multiple of 3 and 9, which is 9:

$$9\left(-\frac{14}{3}\right) + 9\left(\frac{4b}{9}\right) - 9(2) = 9(0)$$ $$-3 \cdot 14 + 4b - 18 = 0$$ $$-42 + 4b - 18 = 0$$

Combine the constant terms:

$$4b - 60 = 0$$

Add 60 to both sides:

$$4b = 60$$

Divide by 4:

$$b = \frac{60}{4}$$ $$b = 15$$

The value of $b$ is $\boxed{15}$.