For each positive integer $n$, let $a(n)$ be the number of zeroes in the base 3 representation of $n$. For which positive real numbers $x$ does the series\[
\sum_{n=1}^\infty \frac{x^{a(n)}}{n^3}
\]converge?
1. **Rewriting the Series**: We start with the series 
   \[
   \sum_{n=1}^\infty \frac{x^{a(n)}}{n^3}
   \]
   where \(a(n)\) is the number of zeroes in the base 3 representation of \(n\).

2. **Grouping by Powers of 3**: To simplify the analysis, we group the terms by powers of 3. Specifically, for \(3^m \le n < 3^{m+1}\), we approximate \(n^3\) by \(27^m\). This approximation is valid because \(3^m \le n < 3^{m+1}\) implies \(27^m \le n^3 < 27^{m+1}\).

3. **Rewriting the Sum**: Using this approximation, we rewrite the series as:
   \[
   \sum_{m=0}^{\infty} \frac{\sum_{3^m \le n < 3^{m+1}} x^{a(n)}}{27^m}
   \]

4. **Counting Zeroes in Base 3**: Next, we need to count the number of zeroes in the base 3 representation of numbers in the range \(3^m \le n < 3^{m+1}\). The number of such numbers with exactly \(k\) zeroes is given by the binomial coefficient \(\binom{m}{k}\) times \(2^{m-k}\) (since each of the remaining \(m-k\) digits can be either 1 or 2).

5. **Summing Over Zeroes**: Therefore, the inner sum can be written as:
   \[
   \sum_{3^m \le n < 3^{m+1}} x^{a(n)} = \sum_{k=0}^m \binom{m}{k} \cdot 2^{m-k} \cdot x^k
   \]

6. **Simplifying the Inner Sum**: This sum is a binomial expansion:
   \[
   \sum_{k=0}^m \binom{m}{k} \left(\frac{x}{2}\right)^k \cdot 2^m = 2^m \left(1 + \frac{x}{2}\right)^m
   \]

7. **Substituting Back**: Substituting this back into our series, we get:
   \[
   \sum_{m=0}^{\infty} \frac{2^m \left(1 + \frac{x}{2}\right)^m}{27^m} = \sum_{m=0}^{\infty} \left(\frac{2}{27}\right)^m \left(1 + \frac{x}{2}\right)^m
   \]

8. **Combining Terms**: This can be further simplified to:
   \[
   \sum_{m=0}^{\infty} \left(\frac{2+x}{27}\right)^m
   \]

9. **Convergence of the Geometric Series**: The series \(\sum_{m=0}^{\infty} r^m\) converges if and only if \(|r| < 1\). Therefore, we need:
   \[
   \left|\frac{2+x}{27}\right| < 1
   \]

10. **Solving the Inequality**: Solving this inequality, we get:
    \[
    \frac{2+x}{27} < 1 \implies 2 + x < 27 \implies x < 25
    \]

Conclusion:
The series converges if and only if \(x < 25\).

The final answer is \( \boxed{ x < 25 } \).