I will begin the calculation, and you can complete it. The random variable $G$ (the number of girls) can take on no values other than $0$, $1$, $2$, $3$, and $4$. The easiest thing to calculate is $P(G=4)$.
How can there be $4$ girls? We must be dealing with a family of $4$ children (probability $0.1$) and that family must have $4$ girls. The probability of $4$ girls given that there are $4$ children is $1/2^4$. So the probability that a family has $4$ children and they are all girls is $(0.1)(1/16)$. To prepare for later, let's do this again with symbols. $P((T=4)\cap (G=4))=P((G=4)|(T=4))P(T=4)=(1/2^4)(0.1)$ Now let us calculate the probability that $G=3$. This event can happen in two ways: (a) The family has $3$ children and they are all girls OR (b) the family has $4$ children and exactly $3$ of them are girls.
First tackle (a). Let us find the probability of $3$ children, all girls. The probability of $3$ children is $0.2$. Given that a family has $3$ children, the probability they are all girls is $1/8$. So the probability of $3$ children and all girls is $(0.2)(1/8)$. Now tackle (b). The probability of $4$ children is $(0.1)$. Given that there are $4$ children, the probability there are exactly $3$ girls is $4/16$ (I hope this part is known to you). So the probability of $4$ children and exactly $3$ girls is $(0.1)(4/16)$. Add up the two probabilities we obtained. The probability of $3$ girls is $(0.2)(1/8)+(0.1)(4/16)$. In symbols, $P(G=3)= P((G=3)|(T=3))P(T=3)+P((G=3)|(T=4))P(T=4)$
Now you can proceed to find the probability of $2$ girls ($3$ cases) also $1$ girl, also $0$ girls. If you have calculated $4$ probabilities, the fifth can be found by subtracting the sum of the others from $1$. But you might as well calculate all $5$ probabilities, and use the fact they must add up to $1$ as a check of your calculations.