Answer
$2.82KHz$
Work Step by Step
We know that
$f^{\prime}=f+\frac{fu}{c}$
This can be rearranged as:
$f^{\prime}-f=\frac{fu}{c}$
We plug in the known values to obtain:
$f^{\prime}-f=\frac{(1.525\times 10^9)(9.00mph)}{3.00\times 10^8}(\frac{0.447m/s}{mph})$
$f^{\prime}-f=1410Hz$
Now, $f^{\prime \prime}=2(f^{\prime}-f)=2(1410)=2.82KHz$
