For an amplitude modulated wave the minimum amplitude is 3 V, while the modulation index is 60%. The maximum amplitude of the modulated wave is:

For an amplitude modulated wave:

$$A_{max} = A_c + A_m$$ and $$A_{min} = A_c - A_m$$

where $$A_c$$ is the carrier amplitude and $$A_m$$ is the modulating signal amplitude.

The modulation index is:

$$ m = \frac{A_m}{A_c} = 0.60 $$

Given $$A_{min} = 3$$ V:

$$ A_c - A_m = 3 $$

$$ A_c - 0.6A_c = 3 $$

$$ 0.4A_c = 3 $$

$$ A_c = 7.5\;\text{V} $$

Therefore $$A_m = 0.6 \times 7.5 = 4.5$$ V.

The maximum amplitude is:

$$ A_{max} = A_c + A_m = 7.5 + 4.5 = 12\;\text{V} $$

The correct answer is 12 V.