D

Step 1: Consider the charges on both spheres. The first sphere has a charge of q and the second sphere has a charge of -2q.
Step 2: When the spheres are connected by a conducting wire, charge will flow until the potential on both spheres is equal.
Step 3: The total charge before connecting the spheres is q - 2q = -q.
Step 4: The combined radius of the two spheres is R for the first sphere and 2R for the second sphere. The potential on a sphere is given by V = \frac{kQ}{R} where Q is the charge and R is the radius.
Step 5: Setting potentials equal for equilibrium: \( \frac{kq}{R} = \frac{k(-2q)}{2R} \).
Step 6: This simplifies to \( \frac{q}{R} = \frac{-2q}{2R} \), leading to |V| being equal.
Step 7: Using charge conservation, the final charge on the first sphere when both potentials are equal can be found.
The final charges will adjust to \( \frac{-q}{3} \) and \( \frac{2q}{3} \) respectively.
Step 8: Thus, charge flowed is final - initial = \( \left( \frac{-q}{3} - q \right) = -\frac{4q}{3} \).
Therefore, upon rewriting in relation to the amount of charge that has flown, we derive the answer as \( \frac{2q}{3} \).
The correct option is D: \( \frac{2q}{3} \).