Two long conductors, separated by a distance d carry currents I₁ and I₂ in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3 times. The new value of the force between them is:
A) -2F
B) F/3
C) -F/3
D) 2F/3

The force between two parallel current-carrying conductors is given by:

\[
F = \frac{\mu_0 I_1 I_2}{2\pi d}
\]

Where:
- \( F \) is the force per unit length between the two conductors.
- \( \mu_0 \) is the permeability of free space.
- \( I_1 \) and \( I_2 \) are the currents in the conductors.
- \( d \) is the distance between the conductors.

### Initial Scenario:
- Currents: \( I_1 \) and \( I_2 \) (same direction)
- Distance: \( d \)
- Force: \( F \)

### New Scenario:
- One current is doubled: \( I_1 \) becomes \( 2I_1 \).
- Its direction is reversed, so it becomes \( -2I_1 \).
- Distance is increased to \( 3d \).

The new force \( F' \) between the conductors is given by:

\[
F' = \frac{\mu_0 (-2I_1) I_2}{2\pi (3d)} = -\frac{2\mu_0 I_1 I_2}{6\pi d} = -\frac{1}{3} \cdot \frac{\mu_0 I_1 I_2}{2\pi d} = -\frac{F}{3}
\]

Thus, the new force is:

\[
F' = -\frac{F}{3}
\]

### Answer:
The correct option is **D. -\dfrac{F}{3}**.