(A) should be the correct choice.
In the community we know that each set of parents will have exactly $1$ boy.
The number of girls might differ.
To find the ratio of boys to girls in the community, we are going to find the the expected number of girls that each parent set can have.
Henceforth in this question we are going to use the word "family" to denote a "parent set".
Calculation of expected number of girls in any family
Let $X$ be a random variable that denotes the number of girls that any family. Each family will have exactly $1$ boy.
$P\left( X = 0\right)$ will denote : Probability that a family has $0$ girls and $1$ boy.
$P\left( X = 0\right) = \left( 0.49\right)^{0}\left( 0.51\right) $
$P\left( X = 3\right)$ will denote : Probability that a family has $3$ girls and $1$ boy.
$P\left( X = 3\right) = \left( 0.49\right)^{3}\left( 0.51\right)$
and so on.
In general we can say that
$P\left( X = i\right)$ will denote the probability that a family has $i$ girls and $1$ boy, and
$P\left( X = i\right) = \left( 0.49\right)^{i}\left( 0.51\right)$ .
Now the expected number of girls in any family will be denoted by $E\left[X\right]$.
Here,
$E\left[X\right] = \sum_{i=0}^{\infty} i\cdot P\left( X = i \right)$.
but $P\left( X = i\right) = \left( 0.49\right)^{i}\left( 0.51\right)$. So, we get,
$E\left[X\right] = \sum_{i=0}^{\infty} i\cdot \left( 0.49\right)^{i}\left( 0.51\right)$.
This implies $E\left[X\right] = \left( 0.51\right)\sum_{i=0}^{\infty} i\cdot \left( 0.49\right)^{i}$.
The formula for summation of series of type $\sum_{k=0}^{\infty} k\cdot x^{k}$ can be found by differentiating
$\sum_{k = 0}^{\infty}x^k$ with respect to $x$.(See the reference below).
This gives
$\sum_{k=0}^{\infty} k\cdot x^{k} = \frac{x}{\left(x-1\right)^2}$
So, $\sum_{i=0}^{\infty} i\cdot \left(0.49\right)^{i} = \frac{0.49}{\left(0.49-1\right)^2}$
Hence,
$E\left[X\right] = \left( 0.51\right)\cdot \frac{0.49}{\left(0.51\right)^2} = \frac{0.49}{0.51}$
Now the ratio of boys to girls can be given by number of boys in each family/expected number of girls in each family.
i.e., $Ratio\left(\text{B to G}\right) = \frac{1}{E\left[X\right]}$
So, $Ratio\left(\text{B to G}\right) = \frac{1}{\frac{0.49}{0.51}} = \frac{51}{49}$
Reference for series summation:
http://math.stackexchange.com/questions/629589/converge-of-the-sum-sum-k-1n-k-xk