Let \( V_1(t) \) be the volume of liquid in the first vat at time \( t \) (in minutes), where it starts with 0 gallons and is filling at a rate of 333 gallons per minute. The volume of the first vat at time \( t \) is given by:
\[ V_1(t) = 333t \]
The first vat has a capacity of 202020 gallons. Therefore, the volume reaches its maximum when:
\[ 333t = 202020 \implies t = \frac{202020}{333} \approx 606.06 \text{ minutes} \]
At that point, the first vat will be at its full capacity.
Now, let \( V_2(t) \) be the volume of liquid in the second vat at time \( t \). The second vat starts with 116116116 gallons and is leaking at a rate of 222 gallons per minute, so we have:
\[ V_2(t) = 116116116 - 222t \]
Next, we need to find the time \( t \) when both vats have the same volume of liquid:
\[ V_1(t) = V_2(t) \]
Substituting the equations we have:
\[ 333t = 116116116 - 222t \]
Combining terms:
\[ 333t + 222t = 116116116 \]
\[ 555t = 116116116 \]
Solving for \( t \):
\[ t = \frac{116116116}{555} \approx 209,839.29 \text{ minutes} \]
Now, to convert this to seconds:
\[ t = 209839.29 \times 60 \approx 12591057.4 \text{ seconds} \]
Thus, the time at which both vats will have the same amount of liquid is approximately \( \boxed{12591057.4} \) seconds.