I know that Pascal's Triangle is constructed by adding the two elements above each element, and that the first and last elements of each row are 1. I also know that the sum of the elements in the $n$th row is $2^n$, since each element is the number of ways to choose a subset of size $k$ from a set of size $n$, and there are $2^n$ possible subsets. Therefore, the arithmetic mean of the elements in the $n$th row is $\frac{2^n}{n+1}$, since there are $n+1$ elements in the row. I need to find the value of $n$ such that $\frac{2^n}{n+1} = 51.2$. This looks like a hard equation to solve algebraically, so I'll try to find a solution by trial and error. I'll start by plugging in some values of $n$ and see how close they get to 51.2. If $n=5$, then $\frac{2^5}{5+1} = \frac{32}{6} \approx 5.3$, which is too small. If $n=6$, then $\frac{2^6}{6+1} = \frac{64}{7} \approx 9.1$, which is still too small. If $n=7$, then $\frac{2^7}{7+1} = \frac{128}{8} = 16$, which is getting closer but not there yet. If $n=8$, then $\frac{2^8}{8+1} = \frac{256}{9} \approx 28.4$, which is still not enough. If $n=9$, then $\frac{2^9}{9+1} = \frac{512}{10} = 51.2$, which is exactly what I want. So, the value of $n$ is 9. 
In the $n$th row of Pascal's Triangle where the first row is $n=0$, the arithmetic mean of the elements is $51.2$. What is the value of $n$?