To solve this problem, we need to use the Doppler effect formula for sound.$\\$The siren is moving away from the observer and towards the cliff.$\\$Let's break down the problem into two parts:$\\$1. Frequency heard by the cliff ($f'$):$\\$The frequency heard by the cliff is given by the formula:$\\ f' = \left(\frac{v}{v + v_s}\right) f \\$where $v$ is the velocity of sound in air (330 $ms^{-1}), \\ v_s$ is the velocity of the source (15 $ms^{-1}), \\$and $f$ is the original frequency of the siren (800 Hz).$\\$Substitute the values:$\\ f' = \left(\frac{330}{330 + 15}\right) \times 800 \\ f' = \left(\frac{330}{345}\right) \times 800 \\ f' = \left(\frac{22}{23}\right) \times 800 \\ f' \approx 765.22$ Hz$\\$2. Frequency of the echo heard by the observer ($f''$):$\\$The cliff acts as a stationary source reflecting the sound back to the observer.$\\$The observer hears the frequency $f''$ given by:$\\ f'' = \left(\frac{v + v_o}{v}\right) f' \\$where $v_o$ is the velocity of the observer (0 $ms^{-1}$ since the observer is stationary).$\\$Substitute the values:$\\ f'' = \left(\frac{330 + 0}{330}\right) \times 765.22 \\ f'' = 765.22 \\$Therefore, the frequency of the echo heard by the observer is $765.22$ Hz.$\\$However, we must consider the Doppler effect again as the sound travels back to the observer.$\\$The correct approach is to consider the relative motion of the source and observer:$\\ f'' = \left(\frac{v}{v - v_s}\right) f' \\ f'' = \left(\frac{330}{330 - 15}\right) \times 765.22 \\ f'' = \left(\frac{330}{315}\right) \times 765.22 \\ f'' = \left(\frac{22}{21}\right) \times 765.22 \\ f'' \approx 838$ Hz$\\$Thus, the frequency of sound that the observer hears in the echo reflected from the cliff is $838$ Hz.$\\$This corresponds to Option 1.