Answer
For an observer on the Earth, the time between ticks is an extra $0.36~ns$
Work Step by Step
Let $\Delta t_0$ be 1 second in the clock's frame.
Let $\Delta t$ be the time measured in the Earth frame.
We can find the time measured in the Earth frame:
$\Delta t = \frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$
$\Delta t \approx (\Delta t_0)(1+\frac{v^2}{2c^2})$
$\Delta t \approx (1~s)\left(1+\frac{(8000~m/s)^2}{(2)(3.0\times 10^8~m/s)^2}~\right)$
$\Delta t = 1.00000000036~s$
For an observer on the Earth, the time between ticks is an extra $0.36~ns$.
