Where blocks would be consecutive 0's or consecutive 1's. So 0000 would be a block of length 4.
I'm not even sure how such a set would look? Would the following elements at least be in the set (so I have a better idea):
0, 1, 01, 10, 000, 111, 0000, 1111, 0011, 1100
For every divisor $d$ of $n$, there are exactly two strings with length $n$ and block length $d$: one starts with $0$'s and the other starts with $1$'s. So if $f(n)$ is the number of binary strings composed of equal block lengths, then $f(n)=2d(n)$ where $d(n)$ counts the number of divisors of $n$.
The ordinary generating function of $f$ is $2\sum_n d(n)x^n$, and you can't say much more than that, though Geoffrey Critzer points out the Lambert series expansion $$\sum_{n\ge1} d(n)x^n = \sum_{n\ge1}\sum_{d|n}x^n=\sum_{d\ge1}\sum_{k\ge1}x^{kd}=\sum_{d\ge1}\frac{x^d}{1-x^d}. $$ However, the Dirichlet generating function is more interesting: $$\sum_{n>0} \frac{f(n)}{n^s} = 2\sum_{n>0}\frac{d(n)}{n^s}=2\zeta^2(s)$$ where $\zeta(s)$ is the Riemann zeta function. (See this link.)