Part 1
Unless I am grossly misinterpreting or missing something very obvious (in which case I obviously welcome any correction), I wouldn't agree that the correct answer is (c). In fact, I wouldn't even agree that the question is well defined. In my opinion, if the question was presented exactly like this at the exam and without further info, then it is unsuitable and ill-posed. For an explanation of that, please see Part 2.
For starters, using the probabilities shown further below, we calculate that $$\mu = \mathbb{E}[X] = \sum_x x\mathbb{P}[X=x] = 1.25,$$ $$\sigma^2 = \mathbb{V}[X] = \left(\sum_{x}x^2\mathbb{P}[X=x]\right) - \mu^2 = 0.6875$$ Therefore, if $Y = 50X$, then $\mu_Y = \mathbb{E}[50X] = 50\mu = 50\cdot 1.25 = 62.5$, while $\sigma_Y^2 = \mathbb{V}[50X] = 50^2\sigma^2 = 50^2\cdot 0.6875 = 1,718.75$.
You have successfully calculated that $\mu_Y = 62.5$, so all that's left is "matching" it to one of the possible values of (c) and (d). As $\sqrt{1,718.15}\approx 41.45$, you can see that the most likely answer is (d), assuming that the question specifies a normal distribution in terms of mean and standard deviation, i.e. $\mathcal{N}(\mu,\sigma)$ instead of mean and variance, i.e. $\mathcal{N}(\mu, \sigma^2)$ (this should have been mentioned in the lecture notes or the description of the question on the exam).
Your confusion about why the final distribution is not binomial but normal is a bit misplaced. The final distribution is binomial, however the binomial distribution is impractical to calculate when $n$ is large. As it happens, in those cases of a large $n$, the binomial distribution with parameters $(n, p)$ can be approximated by a normal distribution with mean $np$ and variance $np(1-p)$, and the larger the $n$ the better the
approximation. For some more details on this, check out the Wiki page.
Part 2
The reason I said the most likely (but not correct) answer is (d) because the distribution of the amount of girls, $X$ is not binomially distributed, because you don't know beforehand what the $n$ parameter will be, i.e. after how many trials the experiment will end. Let's look at what the sample space of three children looks like, depending on what the first child is:
$$\newcommand\pad[1]{\hspace{-18pt}\llap{#1}\phantom{1764}} \hspace{18pt}\begin{matrix} &&&&&&\pad{B}\\ &&&&&\pad{B}&\pad{}&\pad{G}\\ &&&&\pad{B}&\pad{}&\pad{G}&\pad{}&\pad{G}
\end{matrix} \hspace{18pt}\begin{matrix} &&&&&&\pad{G}\\ &&&&&\pad{G}&\pad{}&\pad{B}\\ &&&&\pad{G}&\pad{}&\pad{B}&\pad{}&\pad{G}
\end{matrix}
$$
As it can be seen from here, we can calculate the probabilities of $\{X=x\}$ as follows:
$\mathbb{P}[X = 0] = \mathbb{P}[\text{3 kids, all boys}] = 0.125, $
$\mathbb{P}[X = 1] = \mathbb{P}[\text{2 kids, 1 girl}] + \mathbb{P}[\text{3 kids, 1 girl}]= 0.625, $
$\mathbb{P}[X = 2] = \mathbb{P}[\text{3 kids, 2 girls, in the order BBG}] = 0.125, $
$\mathbb{P}[X = 3] = \mathbb{P}[\text{3 kids, all girls}] = 0.125 $
You calculated these probabilities yourself, of course, as you showed in the comment, but the reason I wrote the calculations nonetheless is because as I mentioned above, we do not know beforehand the number of trials it will take to achieve the desired result, but there is also an upper bound of the amount of trials you can have (in this case, three trials). I would therefore classify this as some kind of bounded negative binomial distribution problem, with an example of the unbounded case seen here.
