Answer
$$608\;Hz$$
Work Step by Step
Let $v$ be the speed of the sound in air. $v_s$ and $v_l$ are the speed of the source and the listener respectively. If $u$ be the speed of the wind, which is blowing toward the whistle (source) and away from the listener, the frequency of the whistle heard by the listener is
$f^{'}=\frac{v+v_l-u}{v-v_s+u}f$
Substituting the given values
$f^{'}=\frac{343+(30.5-30.5)}{343-(30.5+30.5)}\times500\;Hz$
or, $f^{'}=\frac{343}{282}\times500\;Hz$
or, $f^{'}\approx608\;Hz$
The frequency of the whistle heard by the listener is $608\;Hz$
