Since the probability of having a child of either particular gender is 1/2 and is independent, the percentage must be expected to remain constant at 1/2 boys and 1/2 girls. (Anytime anyone has a child we expect it to be a boy with probability 1/2, when they stop having children does not matter.)
Maybe it is illuminating to consider the first couple of years:
At the end of the first year, we expect 1/2 of the couples to have had boys and 1/2 to have had girls, so we have 1/2 of the children are boys, and 1/2 are girls (or 50,000 boys and 50,000 girls using the N = 100,000 couples of the problem).
At the end of the second year, all of the couples with girls will have another child, 1/2 of them expected to be boys, and 1/2 expected to be girls. Hence, the overall population of children will still be 1/2 boys and 1/2 girls. (So, in particular with N = 100,000, we expect 50,000 of these to have already had a boy, and 50,000 to have another child. Of these 50,000 we expect 25,000 to have boys and 25,000 to have girls. Hence, the population of children at the end of the second year is 75,000 boys and 75,000 girls.)
This will continue until every couple has a boy... The child population from the previous year will be 1/2 boys and 1/2 girls, and the new children born in a particular year will be 1/2 boys and 1/2 girls, leaving the ratio of boys to girls unchanged.
This post consists of two parts: a long introduction, and a short solution.
The problem is not optimally worded. We make an interpretation, partly based on the answer provided.
Each family keeps breeding until it has achieved its goal of $2$ girls and then stops, whether or not the other families have managed to meet their quota. And we are invited to assume that the probability that a birth results in a girl is $\dfrac{1}{2}$, and to make the usual assumption of independence.
Introduction: There are a couple of unfortunately different descriptions of the negative binomial distribution. We either count the total number of trials until the $r$-th time that a certain event, often called a "success," occurs. Or else we count the number of what are usually called "successes" until the $r$-th failure. Note that the notion of success and failure are reversed, and even after we do the reversal, the answers differ.
We use the second interpretation of "negative binomial." I checked, this is the interpretation described in Wikipedia. The probability of success is often called $p$. In our case, we end up calling the birth of a girl a failure. Sorry about that!
Let $X_1,X_2,\dots,X_m$ be independent negative binomials, where $X_i$ measures the number of successes until the $r_i$-th failure. Suppose that for each of these $X_i$, the probability of success is $p$. Let $Y=X_1+X_2+\cdots +X_m$. Then $Y$ has negative binomial distribution, with the same "$p$," and $r=r_1+r_2+\cdots+r_n$.
One can prove this by a calculation (the case $m=2$ is enough). But it is also intuitively clear. Because it is important to think the right way about these things, we give a brief explanation.
Take a general negative binomial $W$, the number of successes until the $r$-th failure. Then $W$ is a sum $H_1+H_2+\cdots +H_r$ of $r$ independent random variables, where $H_j$ is the number of successes until the first failure.
Thus $X_1+X_2+\cdots+X_m$ is a sum of $r_1+r_2+\cdots+r_m$ independent random variables of type "$H$," and is therefore negative binomial.
Solution of the problems: So the total number of successes is the number of boys until the $6$-th girl. All of the analysis in the introduction was probably not necessary. The situation is the same as if a single family kept breeding until it had $6$ girls. But the problem is a good excuse for discussing the general situation.
The probability that the total number of boys is $k$ is
$$\binom{k+5}{k}p^k(1-p)^6,$$
where $p$ is the probability of a boy, in this case $\dfrac{1}{2}$.
The mean total number of boys is $\dfrac{6p}{1-p}$, in this case $6$.
Best Answer
The problem defines $X$ as the number of children. Then in part a) it asks about male children. Potentially confusing!
a) There are $x$ male children if there is one female among the first $x+1$ children, and a female on the $(x+1)$-th birth. So the probability that the number of male children is $x$ is equal to $\binom{x+1}{1}\frac{1}{2^{x+2}}$.
b) This is a little different, for it asks for the probability distribution of $X$, the number of children. The same reasoning as the one above shows that $$\Pr(X=x)=\binom{x-1}{1}\frac{1}{2^x}.\tag{1}$$
c) For the probability of at most $4$ children, we want $\Pr(X=2)+\Pr(X=2)+\Pr(X=4)$. Now we can use (1).
Alternately, we could imagine that instead of stopping when there are $2$ females, the person keeps having children until there are $4$. Then the probability for c) is the probability of $2$ or more females in $4$ births. This is a binomial distribution problem, so the probability for c) can also be written as $\frac{\binom{4}{2}+\binom{4}{3}+\binom{4}{4}}{2^4}$.
d) One can look up the expectation of the negative binomial. There is a potential for confusion, because negative binomial is used in two ways, number of successes until the $r$-th failure, or number of trials until the $r$-th success.
Or else one can derive a formula for the expectation, either in general or for our particular case. Attacked directly, that involves summing an infinite series.
Or else we can say that the expected number of children until the second female is the number of children until the first female, plus the number of births from the first female to the second.
The number of children up to and including the first female has geometric distribution, parameter $p=1/2$. Thus the expected number up to and including the first female is $1/p$, which is $2$. Similarly, the number of children between the first female (not inclusive) and the second (inclusive) has geometric distribution with mean $2$.
Thus the mean number of children, as you conjectured, is $4$.