Determine the conditional probability mass function of the size of a randomly chosen family containing 2 girls. Suppose that 15 percent of the families in a certain community have no children, 20 percent have 1, 35 percent have 2, and 30 percent have 3 children; suppose further that each child is equally likely (and independently) to be a boy or a girl. If a family is chosen at random from this community, then B, the number of boys, and G, the number of girls, determine the conditional probability mass function of the size of a randomly chosen family containing 2 girls.
My attempt
There are exactly three ways this can happen:
1) family has exactly 2 girls
  2) family has 2 girls and 1 boy
  3) family has all 3 girls  
The first one is pretty simple. Given that you are going to "select" exactly two children, find the probability that they are BOTH girls (it's a coin flip, so p = 50% = 0.5):
$0.5^2 = 0.25$
So the probability that the family has exactly 2 girls is the probability that the family has exactly two children times the probability that those two children will be girls:
$\frac{1}{4} \cdot 35\% = 8.75\%$
Now find the probability that, given the family has exactly 3 children, that exacly two are girls. Now you flip 3 times but only need to "win" twice-this is a binomial experiment.
There are 3 choose 2 = 3 ways to have exactly two girls: 1st, 2nd, or 3rd is a boy... interestingly the probability of having any particular permutation is just $0.5^3 = 1/8$ (because it's still $0.5 \times 0.5$ for two girls, then $0.5$ for one boy).
So the chance of exactly 2 girls is: $\frac{3}{8}$
Now find the probability for having exactly 3 girls... that's easy, there's only one way, you just have all 3 girls, probability is just $\frac{1}{8}$.
Now, add these up
$\frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$
So now use the percent of families with exactly 3 children to find this portion of the probability:
$\frac{1}{2} \cdot 30\%  = 15\%$
Hence, add the two probabilities... here is it in full detail
$$\begin{eqnarray}\mathbb{P}(\text{contains 2 girls}) &=& \mathbb{P}(\text{2 children}) \times \mathbb{P}(\text{2 girls, 2 children}) + \\ &\phantom{+=}& \mathbb{P}(\text{3 children}) \times \mathbb{P}(\text{2 or 3 girls, 3 children}) \end{eqnarray}$$
$\frac{1}{4} 35\%  + 30\% \times \left(\frac{3}{8} +\frac{ 1}{8}\right)$
$8.75\% + 15\% = 23.75\%$
Is my attempt correct?

The problem statement is somewhat ambiguous. Either we want to find (i) the (conditional) probability distribution of the family size, given that there are at least $2$ girls, or (ii) the (conditional) probability distribution of the family size, given that there are exactly $2$ girls.  The phrase "containing $2$ girls" can be interpreted either way, with some tilting towards (i) as the more likely.
We calculate the probabilities for (i). The probability that there are at least $2$ girls is $(0.35)(1/2)^2 +(0.30)(1/2)$.  The second part of the sum is because if the family has $3$ children, the probability that girls will outnumber boys is, by symmetry, equal to $1/2$. You computed this probability correctly.
That is not what you are being asked to find, though it is a very useful part of the process. Given that there are at least $2$ girls, the family size is either $2$ or $3$. We want to find the conditional probability that the family size is $2$. (Then we will be able to find quickly the conditional probability that the family size is $3$.)
Let $G$ be the event "there are at least $2$ girls" and let $S_2$ be the event "the family size is $2$." We want $P(S_2|G)$. 
By a formula that may be familiar, we have
$$P(S_2|G)P(G)=P(S_2\cap G)=P(G|S_2)P(S_2).$$
Now we are essentially finished. We know $P(G)$, because we computed it above. We know $P(G|S_2)$, it is $(1/2)^2$. And we know $P(S_2)$, we were told what it is in the statement of the problem.
To solve the problem under the assumption (ii), the procedure is almost the same, but the probabilities are different.  For example, the probability that a family has exactly $2$ girls is $(0.35)(1/2)^2+(0.30)(3/8)$.  
