As we know that the intensity of sound waves is related to distance

\(\displaystyle{I}\alpha{\frac{{{1}}}{{{r}^{{{2}}}}}}\)

where I is the intensity of light

r is the distance between the source and the observer

so \(\displaystyle{\frac{{{I}_{{{1}}}}}{{{I}_{{{2}}}}}}={\frac{{{{r}_{{{2}}}^{{{2}}}}}}{{{{r}_{{{1}}}^{{{2}}}}}}}\)

we have \(\displaystyle{I}_{{{1}}}={I}\)

\(\displaystyle{I}_{{{2}}}={0.25}{I}\)

\(\displaystyle{r}_{{{1}}}={1450}{m}\)

substituting these values we get

\(\displaystyle{r}_{{{2}}}={2900}{m}\)

hence the helicopter must fly 2900m above the ground

\(\displaystyle{I}\alpha{\frac{{{1}}}{{{r}^{{{2}}}}}}\)

where I is the intensity of light

r is the distance between the source and the observer

so \(\displaystyle{\frac{{{I}_{{{1}}}}}{{{I}_{{{2}}}}}}={\frac{{{{r}_{{{2}}}^{{{2}}}}}}{{{{r}_{{{1}}}^{{{2}}}}}}}\)

we have \(\displaystyle{I}_{{{1}}}={I}\)

\(\displaystyle{I}_{{{2}}}={0.25}{I}\)

\(\displaystyle{r}_{{{1}}}={1450}{m}\)

substituting these values we get

\(\displaystyle{r}_{{{2}}}={2900}{m}\)

hence the helicopter must fly 2900m above the ground