First, find the coefficient of \(x^2\) in \(\left( 2 + \frac{x}{2} \right)^6\).

Using the binomial theorem, the term containing \(x^2\) is given by:

\(\binom{6}{2} \cdot 2^4 \cdot \left( \frac{x}{2} \right)^2 = 15 \cdot 16 \cdot \frac{x^2}{4} = 60x^2\)

So, the coefficient of \(x^2\) in \(\left( 2 + \frac{x}{2} \right)^6\) is 60.

Next, find the coefficient of \(x^2\) in \((a + x)^5\).

Using the binomial theorem, the term containing \(x^2\) is given by:

\(\binom{5}{2} \cdot a^3 \cdot x^2 = 10a^3x^2\)

So, the coefficient of \(x^2\) in \((a + x)^5\) is \(10a^3\).

According to the problem, the sum of these coefficients is 330:

\(60 + 10a^3 = 330\)

Solve for \(a^3\):

\(10a^3 = 270\)

\(a^3 = 27\)

\(a = 3\)