There are initially 18 balls, and as you mentioned, once you have chosen and removed the first two (white) balls, you are left with 16 balls.
Of those 16 balls, 6 are green. So the probability of picking a green ball is 6/16.
All of the boxes contain $N - 1$ balls. This is just a complicated conditional probability problem. Lets look at a single box with $r$ red balls and $g$ green balls. What would the probability be of getting green on the second? Well it depends on whether or not you draw a red or green first. If you draw a red first, then there are $\left.p(\text{second green } \right| \text{ first red}) = \frac{g}{r + g - 1}$. However, if you draw a green ball first then you have one less green to choose from giving: $\left.p(\text{second green } \right| \text{ first green}) = \frac{g - 1}{g + r - 1}$. So what are the chances of each condition happening? $p(\text{first red}) = \frac{r}{g + r}$ and $p(\text{first green}) = \frac{g}{r + g}$. Therefore we can finally write:
\begin{align}
p(\text{second green}) =& \left.p(\text{second green } \right| \text{ first red})p(\text{first red}) + \left.p(\text{second green } \right| \text{ first green})p(\text{first green})\\
=& \frac{r}{r + g}\frac{g}{r+g-1} + \frac{g}{r + g}\frac{g-1}{r+g-1} = \frac{g(r + g - 1)}{(r + g)(r + g - 1)} = \frac{g}{r + g}
\end{align}
Not surprising that drawing the second green has just as good of a chance of being green as the first pick.
Therefore for each of the $N$ boxes you need to compute $p(\text{second green})$ (which is just the probability of drawing a green on the first try). Now the condition is that we choose box $r$ which has $p(\text{second green}) = p(\text{first green}) = \frac{N - r}{N - 1}$. The probability of choosing box $r$ among $N$ boxes is just $\frac{1}{N}$ which gives:
$$
p(\text{second green}) = \sum_1^N \frac{1}{N}\frac{N - r}{N - 1} = \frac{1}{N(N - 1)}\sum_1^r (N - r)
$$
The first sum is very easy (you're just summing the same number, $N$, $N$ times) $\sum_1^N N = N\cdot N = N^2$. The second part is easy if you remember the sum of the first $n$ consecutive integers is $\sum_1^n i = \frac{n(n + 1)}{2}$. So this gives:
$$
p(\text{green}) = \frac{N^2 - \frac{N(N + 1)}{2}}{N(N - 1)} = \frac{2N^2 - N^2 - N}{2N(N - 1)} = \frac{N^2 - N}{2\left(N^2 - N\right)} = \frac{1}{2}
$$
For part $2$), we actually already computed that above: $\left.p(\text{second green }\right|\text{ first green}) = \frac{g - 1}{g + r - 1}$. But now you need to sum over the condition that it could be any of the $N$ boxes (edit: However, the last box, box $N$, has $0$ green balls (and thus seeing green first means it definitely wasn't this box. So we should only sum over the first $N - 1$ boxes and divide by $N - 1$, not $N$.):
\begin{align}
\left.p(\text{second green }\right|\text{ first green}) =& \sum_1^{N - 1}
\frac{1}{N - 1}\frac{N - r - 1}{N - 2} \\
=& \frac{N(N - 1) - (N - 1) - \frac{N(N - 1)}{2}}{N(N - 2)} \\
=& \frac{2N(N - 1) - 2(N - 1) - N(N - 1)}{2(N - 1)(N - 2)}\\
=& \frac{N(N - 1) - 2(N - 1))}{2(N - 1)(N - 2)} \\
=& \frac{(N - 1)(N - 2)}{2(N - 1)(N - 2)} \\
=& \frac{1}{2}
\end{align}
This is only valid for $N > 2$ (since if $N = 1$ there are no balls in each box and if $N = 2$ there is only one ball in each box). This result just confirms that drawing balls are independent events.
Best Answer
Since the colour of the first ball is not known, the probability of the second and third being green is the same as if you only took out two balls and want the probability of them both being green. In both cases, you are essentially choosing $2$ balls at random from the $12$ balls.
This simplifies the problem. The probability of the first green is $\dfrac{7}{12}$. The probability of the second green is then $\dfrac{6}{11}$. Multiplying gives the required probability:
$$\dfrac{7}{12} \dfrac{6}{11} = \dfrac{7}{22}.$$