Answer
The change in frequency heard by the driver is $~-68.8~Hz$
Work Step by Step
We can convert the car's speed to units of $m/s$:
$v = 85~km/h \times \frac{1000~m}{1~km}\times \frac{1~hr}{3600~s} = 23.6~m/s$
We can use the equation for the Doppler effect when the observer is approaching:
$f_i = \left(\frac{v+v_o}{v}\right)~f_s$
$f_i = \left(\frac{343~m/s+23.6~m/s}{343~m/s}\right)~(500~Hz)$
$f_i = 534.4~Hz$
We can use the equation for the Doppler effect when the observer is moving away:
$f_f = \left(\frac{v-v_o}{v}\right)~f_s$
$f_f = \left(\frac{343~m/s-23.6~m/s}{343~m/s}\right)~(500~Hz)$
$f_f = 465.6~Hz$
We can find the change in frequency:
$f_f-f_i = 465.6 - 534.4~Hz = -68.8~Hz$
The change in frequency heard by the driver is $~-68.8~Hz$.
