To find the term independent of x, we use the binomial expansion formula:

\(\left( a + b \right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).

Here, \(a = 2x\) and \(b = \frac{1}{x^2}\), and \(n = 6\).

The general term is:

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

Simplifying, we get:

\(T_k = \binom{6}{k} \cdot 2^{6-k} \cdot x^{6-k} \cdot x^{-2k}\).

For the term to be independent of x, the power of x must be zero:

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

Solving for \(k\), we get:

\(6 - 3k = 0\)

\(3k = 6\)

\(k = 2\).

Substitute \(k = 2\) back into the expression for \(T_k\):

\(T_2 = \binom{6}{2} \cdot 2^{4} \cdot x^{0}\).

\(T_2 = 15 \cdot 16 \cdot 1\).

\(T_2 = 240\).

Thus, the term independent of x is 240.