In amplitude modulation, the message signal $$V_m(t) = 10\sin 2\pi \times 10^5 t$$ volts and carrier signal $$V_C(t) = 20\sin 2\pi \times 10^7 t$$ volts. The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is $$\alpha$$ kHz. The value of $$\alpha$$ is:

We have the message (modulating) signal written as $$V_m(t)=10\sin\bigl(2\pi\times10^{5}\,t\bigr)\;\text{volts}.$$

From the standard trigonometric form $$V_m(t)=V_{m0}\sin(2\pi f_m t),$$ we can directly read the message frequency:

$$2\pi f_m = 2\pi \times 10^{5}\;\Rightarrow\;f_m = 10^{5}\ \text{Hz}.$$

Converting hertz to kilohertz, we obtain

$$f_m = 10^{5}\ \text{Hz}=100\ \text{kHz}.$$

Next, the carrier signal is given as $$V_C(t)=20\sin\bigl(2\pi\times10^{7}\,t\bigr)\;\text{volts}.$$

Again comparing with $$V_C(t)=V_{c0}\sin(2\pi f_c t),$$ we identify

$$2\pi f_c = 2\pi\times10^{7}\;\Rightarrow\;f_c = 10^{7}\ \text{Hz} = 10,000\ \text{kHz}.$$

For amplitude modulation (A.M.), the modulated spectrum consists of three principal components:

1. The carrier at frequency $$f_c.$$

2. The upper sideband (USB) at frequency $$f_c + f_m.$$

3. The lower sideband (LSB) at frequency $$f_c - f_m.$$

The bandwidth $$B$$ of an A.M. signal is defined, and should always be remembered, as

$$B = 2f_m,$$

because the two sidebands lie symmetrically $$f_m$$ hertz above and below the carrier.

Substituting our previously found value $$f_m = 100\ \text{kHz}$$, we obtain:

$$B = 2 \times 100\ \text{kHz} = 200\ \text{kHz}.$$

This bandwidth is denoted by $$\alpha$$ in the statement of the question, so

$$\alpha = 200\ \text{kHz}.$$

Hence, the correct answer is Option A.