The audio signal voltage is given by $V_m=2\sin 12\pi \text{\hspace{0.17em}x\hspace{0.17em}}{10}^{3}t$ . The bandwidth and LSB if carrier waves has a frequency $3.14\text{\hspace{0.17em}x\hspace{0.17em}}{10}^6$  rad/sec

$$\begin{gathered} Given : V_m=A_m\sin w_{m}t \\ here w_m=2\pi f_m=12\pi \text{x}{10}^3 \\ \Rightarrow f_m=6\text{x}{10}^3 = 6 kHz \\ and \\ w_c=3.14\text{\hspace{0.17em}x}{10}^{6}rad/s \\ \Rightarrow 2\pi f_c=3.14\text{\hspace{0.17em}x}{10}^{6}rad/s \\ \Rightarrow f_c=\frac{\pi \text{x}{10}^6}{2\pi }Hz=500 kHz \\ band width = 2f_m=2\text{\hspace{0.17em}x\hspace{0.17em}}6\text{\hspace{0.17em}x\hspace{0.17em}}{10}^3=12\text{\hspace{0.17em}k}Hz \\ LSB=f_c-f_m=500-6=494\text{\hspace{0.17em}k}Hz \end{gathered}$$