$$
begin{align*}
f_{observed}&=f_{source}(1+frac{v_{source}}{c})\
frac{f_{observed}}{f_{source}}&=(1+frac{v_{source}}{c})>1\
f_{observed}&>f_{source}
end{align*}
$$

Therefore when the source is directly moving towards me, the observed
frequency $f_{1}$ is

$$
f_{1}=f_{source}left(1+frac{v_{s}}{c}right)=f_{source}left(1+frac{c/2}{c}right)=frac{3}{2}f_{source}
$$

And when the source is moving directly away from me, the observed
frequency is

$$
f_{2}=f_{source}left(1-frac{v_{s}}{c}right)=f_{source}left(1-frac{c/2}{c}right)=frac{1}{2}f_{source}
$$

So the ratio is

$$
frac{f_{1}}{f_{2}}=3
$$

Speed of light is $c=3times10^{8} {rm m/s}$

Therefore

one light year is

$$
1 {rm ly}=left(3times10^{8} {rm m/s}right)left(3.15times10^{7} {rm s}right)=9.45times10^{15} {rm m}=9.45times10^{12} {rm km}
$$

$f_{s}=5.000times10^{14}$ Hz

$$
v_{s}=3325 {rm km/s}=3325times10^{3} {rm m/s}
$$

Unknown:

$$
f_{o}=?
$$

Solution:

$$
f_{o}=f_{s}left(1-frac{v_{s}}{c}right)=left(5.000times10^{14} {rm Hz}right)left(1-frac{left(3325times10^{3} {rm m/s}right)}{left(3times10^{8} {rm m/s}right)}right)=4.945times10^{14} {rm Hz}
$$