Q.A carrier amplitude to a depth of 50% by a sinusoidal produces side frequency of 335 kHz and 337 kHz. The amplitudes of each side frequency is 15 V. Find the frequency and amplitude of the carrier signal.

The carrier frequency is the mean of the two side frequencies, $f_c = (f_{LSB}+f_{USB})/2$, and the carrier amplitude comes from the side-frequency amplitude relation $V_{SB} = m_a V_C/2$ …

[!TLDR]

Using $V_{SB} = m_a V_C/2$, the carrier amplitude is $V_C = 60$ V; the carrier frequency is the mean of the side frequencies, $f_c = (335+337)/2 = 336$ kHz.

An AM wave produces two side frequencies placed symmetrically about the carrier at $f_c - f_m$ and $f_c + f_m$. The carrier therefore lies exactly midway between them:

Each side frequency has amplitude $V_{SB} = m_a V_C/2$, which can be rearranged to find the carrier amplitude.

Given depth of modulation $m_a = 0.5$, side frequencies 335 kHz and 337 kHz, and side-frequency amplitude 15 V:

$15 = \frac{m_a V_C}{2} = \frac{0.5\,V_C}{2} \implies V_C = \frac{15\times2}{0.5} = 60\ \text{V}$ …