Find the value of real parameter $ a$ such that $ 2$ is the smallest integer solution of \[ \frac{x\plus{}\log_2 (2^x\minus{}3a)}{1\plus{}\log_2 a} >2.\]
To find the value of the real parameter \( a \) such that \( 2 \) is the smallest integer solution of the inequality
\[
\frac{x + \log_2 (2^x - 3a)}{1 + \log_2 a} > 2,
\]
we need to analyze the inequality step by step.

1. **Identify the domain of the expression:**
   - The expression \( \log_2 (2^x - 3a) \) is defined if \( 2^x - 3a > 0 \).
   - Therefore, \( 2^x > 3a \).

2. **Case I: \( 1 + \log_2 a > 0 \Rightarrow a > \frac{1}{2} \):**
   - Rewrite the inequality:
     \[
     x + \log_2 (2^x - 3a) > 2(1 + \log_2 a).
     \]
   - Simplify the right-hand side:
     \[
     2(1 + \log_2 a) = \log_2 (4a^2).
     \]
   - Thus, the inequality becomes:
     \[
     x + \log_2 (2^x - 3a) > \log_2 (4a^2).
     \]
   - Exponentiate both sides to base 2:
     \[
     2^x (2^x - 3a) > 4a^2.
     \]
   - Simplify:
     \[
     2^{2x} - 3a \cdot 2^x > 4a^2.
     \]
   - Rearrange:
     \[
     2^{2x} - 3a \cdot 2^x - 4a^2 > 0.
     \]
   - Factorize:
     \[
     (2^x - 4a)(2^x + a) > 0.
     \]
   - Analyze the factors:
     - Both factors are positive if \( 2^x > 4a \) and \( 2^x > -a \). Since \( a > \frac{1}{2} \), \( 2^x > -a \) is always true.
     - Therefore, \( 2^x > 4a \).

3. **Determine the smallest integer \( x \) such that \( 2^x > 4a \):**
   - For \( x = 2 \):
     \[
     2^2 = 4 \Rightarrow 4 > 4a \Rightarrow a < 1.
     \]
   - For \( x = 1 \):
     \[
     2^1 = 2 \Rightarrow 2 > 4a \Rightarrow a < \frac{1}{2}.
     \]
   - Since \( a > \frac{1}{2} \), \( x = 1 \) does not fit the condition. Therefore, \( x = 2 \) is the smallest integer solution if \( \frac{1}{2} < a < 1 \).

4. **Case II: \( 1 + \log_2 a < 0 \Rightarrow 0 < a < \frac{1}{2} \):**
   - Rewrite the inequality:
     \[
     x + \log_2 (2^x - 3a) < 2(1 + \log_2 a).
     \]
   - Simplify the right-hand side:
     \[
     2(1 + \log_2 a) = \log_2 (4a^2).
     \]
   - Thus, the inequality becomes:
     \[
     x + \log_2 (2^x - 3a) < \log_2 (4a^2).
     \]
   - Exponentiate both sides to base 2:
     \[
     2^x (2^x - 3a) < 4a^2.
     \]
   - Simplify:
     \[
     2^{2x} - 3a \cdot 2^x < 4a^2.
     \]
   - Rearrange:
     \[
     2^{2x} - 3a \cdot 2^x - 4a^2 < 0.
     \]
   - Factorize:
     \[
     (2^x - 4a)(2^x + a) < 0.
     \]
   - Analyze the factors:
     - Since \( 2^x + a > 0 \) and \( 2^x - 4a < 0 \), we must have \( 2^x < 4a \).

5. **Determine the smallest integer \( x \) such that \( 2^x < 4a \):**
   - For \( x = 2 \):
     \[
     2^2 = 4 \Rightarrow 4 < 4a \Rightarrow a > 1.
     \]
   - Since \( 0 < a < \frac{1}{2} \), there are no permissible values of \( a \) in this case.

Therefore, the only values of \( a \) such that the smallest integer solution is \( x = 2 \) is \( a \in \left( \frac{1}{2}, 1 \right) \).

The final answer is \( \boxed{ a \in \left( \frac{1}{2}, 1 \right) } \).