I am trying to find a formula to calculate the staring value of a geometric progression.
Starting value $a$ and ratio $r$. The infinite sum of this sequence is $200$.
Ratio $r$ is calculated: $ 1-(a/200)$
If $a = 52$ then the sum of the first 10 terms: $ar^0 ... ar^9 = 190$ which is 95% of the infinite sum.
The formula for the sum of the first ten terms is: $a\left(\frac{1-r^{10}}{1-r}\right)$
What is the formula to calculate $a$ if I have the infinite sum, the formula for $r$ as above, and the percentage of the infinite sum that I want to arrive at by the 10th term?
I think you made a mistake in calculating $r$. You said $\displaystyle r=1-\frac{S_\infty}{a}$; it should be:
$$\begin{aligned}S_\infty&=\frac{a}{1-r}\\1-r&=\frac{a}{S_\infty}\\r&=1-\frac{a}{S_\infty}\end{aligned}$$
After that, say we want the partial sum of the first $n$ terms to make up proportion $p$ of the infinite series; then:
$$\begin{aligned}p&=\frac{S_{n}}{S_\infty}\\&=\frac{\frac{a\left(1-r^{n}\right)}{1-r}}{\frac{a}{1-r}}\\&=1-r^{n}\\&=1-\left(1-\frac{a}{S_\infty}\right)^{n}\end{aligned}$$
From here, we just need to solve for $a$:
$$\begin{aligned}\left(1-\frac{a}{S_\infty}\right)^{n}&=1-p\\1-\frac{a}{S_\infty}&=\sqrt[n]{1-p}\\a&=S_\infty\left(1-\sqrt[n]{1-p}\right)\end{aligned}$$
In your example, you have $S_\infty=200$, $n=10$ and $p=0.95(=95\%)$, which gives us $a\approx51.8$.