Let $a_0$, $a_1$, $a_2$, $\dots$ be an infinite sequence of real numbers such that $a_0 = \frac{4}{5}$ and
\[
  a_{n} = 2 a_{n-1}^2 - 1
\]
for every positive integer $n$.  Let $c$ be the smallest number such that for every positive integer $n$, the product of the first $n$ terms satisfies the inequality
\[
  a_0 a_1 \dots a_{n - 1} \le \frac{c}{2^n}.
\]
What is the value of $100c$, rounded to the nearest integer?
1. **Initial Setup and Sequence Definition:**
   Given the sequence \(a_0, a_1, a_2, \ldots\) defined by:
   \[
   a_0 = \frac{4}{5}
   \]
   and
   \[
   a_n = 2a_{n-1}^2 - 1 \quad \text{for every positive integer } n.
   \]

2. **Trigonometric Substitution:**
   Let \(\theta = \arccos \left(\frac{4}{5}\right)\). Then:
   \[
   a_0 = \cos \theta = \frac{4}{5}
   \]
   and since \(\sin^2 \theta + \cos^2 \theta = 1\), we have:
   \[
   \sin \theta = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}.
   \]

3. **Recursive Formula in Terms of Cosine:**
   We observe that the recursive formula \(a_n = 2a_{n-1}^2 - 1\) is the double-angle formula for cosine:
   \[
   \cos 2\theta = 2\cos^2 \theta - 1.
   \]
   Therefore, we can express \(a_n\) as:
   \[
   a_n = \cos(2^n \theta).
   \]

4. **Product of Terms:**
   We need to find the product of the first \(n\) terms:
   \[
   \prod_{k=0}^{n-1} a_k = \prod_{k=0}^{n-1} \cos(2^k \theta).
   \]

5. **Telescoping Product Identity:**
   Using the identity for the product of cosines:
   \[
   \prod_{k=0}^{n-1} \cos(2^k \theta) = \frac{\sin(2^n \theta)}{2^n \sin \theta}.
   \]

6. **Bounding the Product:**
   Since \(\sin(2^n \theta)\) can take values as close to 1 as desired, we have:
   \[
   \left| \prod_{k=0}^{n-1} \cos(2^k \theta) \right| \leq \frac{1}{2^n \sin \theta}.
   \]
   Given \(\sin \theta = \frac{3}{5}\), we get:
   \[
   \left| \prod_{k=0}^{n-1} \cos(2^k \theta) \right| \leq \frac{1}{2^n \cdot \frac{3}{5}} = \frac{5}{3 \cdot 2^n}.
   \]

7. **Determining \(c\):**
   Therefore, the smallest \(c\) such that:
   \[
   \prod_{k=0}^{n-1} a_k \leq \frac{c}{2^n}
   \]
   is \(c = \frac{5}{3}\).

8. **Calculating \(100c\):**
   \[
   100c = 100 \cdot \frac{5}{3} = \frac{500}{3} \approx 166.67.
   \]
   Rounded to the nearest integer, we get:
   \[
   100c \approx 167.
   \]

The final answer is \(\boxed{167}\).