We are given the equation \( (23^3)^2 = 4^x \). Let's simplify the left side using the power of a power rule: \( (a^m)^n = a^{mn} \).

\( (23^3)^2 = 23^{3 \times 2} = 23^6 \)

So our equation becomes:

\( 23^6 = 4^x \)

Taking the logarithm (base 10) on both sides:

\( \log(23^6) = \log(4^x) \)

Using the logarithm power rule: \( \log(a^b) = b \log(a) \), we get:

\( 6 \log(23) = x \log(4) \)

Solving for x:

\( x = \frac{6 \log(23)}{\log(4)} \)

This equation is difficult to solve directly for an integer value of x. However, we are given that \( (23^3)^2 = 4^x \). Let's look at this another way.

We have \( 23^6 = 4^x \). We know that \( 4 = 2^2 \). Substituting this, we get \( 23^6 = (2^2)^x = 2^{2x} \). This equation is still difficult to solve for x directly. Let's explore the options to find the correct answer.

Let's try each option to see which one satisfies the given condition. If we assume 3^x=27, then x=3 because 3³=27. Let's check if this x value satisfies the original equation (23³)² = 4^x:

(23³)² = 23⁶ ≈ 148035889

4³ = 64

The values do not match. Let's consider another approach. Note that we have \(23^6 = 4^x\). If we express 4 as 2², we get \(23^6 = (2^2)^x = 2^{2x}\). This doesn't directly help us solve for x easily.

The problem likely contains an error or is designed to test understanding of logarithmic properties without straightforward calculations. Checking the options systematically is the best approach given the provided equation's structure. It is not possible to derive an integer solution for x satisfying the given equation directly, making the options provided the most likely route to answer the question.