Given that $ 2^{2004}$ is a $ 604$-digit number whose ﬁrst digit is $ 1$, how many elements of the set $ S \equal{} \{2^0,2^1,2^2, \ldots,2^{2003}\}$ have a ﬁrst digit of $ 4$? 

$ \textbf{(A)}\ 194 \qquad
\textbf{(B)}\ 195 \qquad
\textbf{(C)}\ 196 \qquad
\textbf{(D)}\ 197 \qquad
\textbf{(E)}\ 198$
1. **Understanding the Problem:**
   We need to determine how many elements in the set \( S = \{2^0, 2^1, 2^2, \ldots, 2^{2003}\} \) have a first digit of 4. Given that \( 2^{2004} \) is a 604-digit number whose first digit is 1, we can use Benford's Law to solve this problem.

2. **Applying Benford's Law:**
   Benford's Law states that the leading digit \( d \) (where \( d \) ranges from 1 to 9) in many naturally occurring collections of numbers is not uniformly distributed. The probability \( P(d) \) that the leading digit is \( d \) is given by:
   \[
   P(d) = \log_{10} \left(1 + \frac{1}{d}\right)
   \]
   For \( d = 4 \):
   \[
   P(4) = \log_{10} \left(1 + \frac{1}{4}\right) = \log_{10} \left(\frac{5}{4}\right)
   \]

3. **Calculating \( \log_{10} \left(\frac{5}{4}\right) \):**
   \[
   \log_{10} \left(\frac{5}{4}\right) = \log_{10} 5 - \log_{10} 4
   \]
   Using the approximations \( \log_{10} 5 \approx 0.6990 \) and \( \log_{10} 4 \approx 0.6021 \):
   \[
   \log_{10} \left(\frac{5}{4}\right) \approx 0.6990 - 0.6021 = 0.0969
   \]

4. **Finding the Number of Elements:**
   Let \( n \) be the number of elements in the set \( S \) that have a first digit of 4. According to Benford's Law:
   \[
   \frac{n}{2004} = \log_{10} \left(\frac{5}{4}\right)
   \]
   Therefore:
   \[
   n = 2004 \times \log_{10} \left(\frac{5}{4}\right) \approx 2004 \times 0.0969 = 194.1876
   \]

5. **Rounding to the Nearest Integer:**
   Since \( n \) must be an integer, we round 194.1876 to the nearest integer, which is 194.

Conclusion:
\[
\boxed{194}
\]