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.