To find the term independent of x in the expansion of \(\left( 2x + \frac{1}{4x^2} \right)^6\), we use the binomial theorem:

\(\binom{6}{k} (2x)^{6-k} \left( \frac{1}{4x^2} \right)^k\)

The term independent of x occurs when the powers of x cancel out, i.e., when:

\((6-k) - 2k = 0\)

Solving for k gives:

\(6 - 3k = 0\)

\(k = 2\)

Substitute \(k = 2\) into the binomial term:

\(\binom{6}{2} (2x)^4 \left( \frac{1}{4x^2} \right)^2\)

\(= 15 \times 16x^4 \times \frac{1}{16x^4}\)

\(= 15\)

Thus, the term independent of x is 15.