To find the coefficients of x² and x³ in the expansion of (3 - 2x)⁶, we use the binomial theorem:

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

For the term involving x², we have:

\(a = \binom{6}{2} \cdot 3^{4} \cdot (-2x)^{2}\)

\(a = 15 \cdot 81 \cdot 4\)

\(a = 4860\)

For the term involving x³, we have:

\(b = \binom{6}{3} \cdot 3^{3} \cdot (-2x)^{3}\)

\(b = 20 \cdot 27 \cdot (-8)\)

\(b = -4320\)

Thus, \(\frac{a}{b} = \frac{4860}{-4320} = -\frac{9}{8}\).

Therefore, the value of \(\frac{a}{b}\) is \(\frac{9}{8}\).