A message signal of frequency 20 kHz and peak voltage of 20 V is used to modulate a carrier wave of frequency 1 MHz and peak voltage of 20 V. The modulation index will be:

We begin by recalling the definition of the modulation index for amplitude modulation. In AM, the modulation index (usually denoted by $$m$$) is the ratio of the peak (maximum) voltage of the message, or modulating, signal $$V_m$$ to the peak voltage of the unmodulated carrier signal $$V_c$$.

Mathematically, the formula is stated as:

$$m=\dfrac{V_m}{V_c}$$

From the data given in the problem we identify:

$$V_m = 20 \text{ V}$$ (because the peak voltage of the message signal is 20 V),

$$V_c = 20 \text{ V}$$ (because the peak voltage of the carrier wave is also 20 V).

Now we substitute these numerical values into the formula:

$$m = \dfrac{V_m}{V_c} = \dfrac{20 \text{ V}}{20 \text{ V}}$$

Dividing the identical numerators and denominators, we get:

$$m = 1$$

Hence, the modulation index is unity, i.e. its value is exactly 1.

So, the answer is $$1$$.