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 $6$.
It's one possibility out of $2^6$ equally likely possibilities. So $\frac 1 {2^6} = \frac 1 {64}$.
The first calculation you did is correct because the probabilities are independent.
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In more detail: $$\Pr[\text{first four children are boys and rest are girls}] = \\\Pr[\text{second to fourth children are boys and rest are girls} \mid \text{first child is a boy}] \cdot \Pr[\text{First child is a boy}] = \\\Pr[\text{second to fourth children are boys and rest are girls}] \cdot \Pr[\text{First child is a boy}]\text{ (by independence)}$$
Best Answer
Assuming the question is what proportion of Russian families with exactly $n$ children will have at least $k$ boys?
The number of boys in a russian family of $n$ kids has a Binomial distribution of parameters $n$ and $p=\frac{1.09}{2.09}$
Which means that they will have exactly $k$ children with probability
$C^k_np^k(1-p)^{n-k}$
Thus the probability of having at least $k$ children is :
$\sum_{i=k}^nC^i_np^i(1-p)^{n-i}=1-\sum_{i=0}^{k-1}C^i_np^i(1-p)^{n-i}$
There is, to my knowledge, no exact simplification of this formula
Well, just replace $n$ with 6 and $k$ with 3.