To find the coefficient of \(x^2\) in the expansion of \(\left( k + \frac{1}{3}x \right)^5\), we use the binomial theorem:

\(\binom{5}{2} k^{5-2} \left( \frac{1}{3}x \right)^2\)

The coefficient of \(x^2\) is:

\(\binom{5}{2} k^3 \left( \frac{1}{3} \right)^2\)

\(= 10 \times k^3 \times \frac{1}{9}\)

We know this coefficient is 30, so:

\(10 \times k^3 \times \frac{1}{9} = 30\)

\(\frac{10}{9} k^3 = 30\)

\(k^3 = 30 \times \frac{9}{10}\)

\(k^3 = 27\)

\(k = 3\)