To find the coefficient of \(x^2\) in the expansion of \(\left( \frac{x}{2} + \frac{2}{x} \right)^6\), we use the binomial theorem.

The general term in the expansion is given by:

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

We need the term where the power of \(x\) is 2:

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

\(6 - 2k = 2\)

\(2k = 4\)

\(k = 2\)

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

\(T_2 = \binom{6}{2} \left( \frac{x}{2} \right)^4 \left( \frac{2}{x} \right)^2\)

\(= 15 \left( \frac{x^4}{16} \right) \left( \frac{4}{x^2} \right)\)

\(= 15 \times \frac{x^4}{16} \times \frac{4}{x^2}\)

\(= 15 \times \frac{x^2}{4}\)

\(= \frac{15}{4} x^2\)

Thus, the coefficient of \(x^2\) is \(\frac{15}{4}\) or 3.75.