Answer
$f=4.86\times10^{14}Hz$.
Work Step by Step
The Doppler effect shifts the frequency of the radiation, due to the star’s motion.
The star is moving away from earth.
$$f=\sqrt{\frac{c-u}{c+u}}f_0$$
$$f=\sqrt{\frac{c-0.520c}{c+0.520c}}(8.64\times10^{14}Hz)$$
$$f=4.86\times10^{14}Hz$$
