Step 1: Understand the question.
Light of frequency twice the threshold falls on a metal and kicks out electrons. Then we drop the frequency to one third of that and double the brightness (intensity). We must say what happens to the current.
Step 2: The one rule that decides everything.
Electrons come out only if the light frequency is at least the threshold frequency $\nu_0$. Below threshold, nothing comes out, no matter how bright the light is.
Step 3: Note the starting frequency.
At first the frequency is $\nu = 2\nu_0$, which is above $\nu_0$, so electrons do come out. Fine so far.
Step 4: Find the new frequency.
The new frequency is one third of $2\nu_0$, that is $\nu' = \frac{1}{3}(2\nu_0) = \frac{2}{3}\nu_0$.
Step 5: Compare with threshold.
Since $\frac{2}{3}\nu_0$ is less than $\nu_0$, the light is now below threshold.
Step 6: Decide the current.
Below threshold no electrons are ejected, so doubling the intensity does nothing. The photoelectric current becomes zero, which is option (C).
\[ \boxed{\text{Current} = 0} \]