To what extent can time be dilated on Earth? Though the value would be pretty small, I was still curious about knowing that numeric value.
 A: Here I am writing the gravitational time dilation, what is caused by the gravity of the Earth.
The spacetime around spherical, not very fast rotating bodies can be described with the Schwarzschild-metric:
$c^2 {d \tau}^{2} =
\left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$
On the wiki link can you read more from it. Here we want to know the proper time distance between two points differing only in the $t$-coordinate, so our formula is significantly simplified to:
${d \tau}^2 = \left(1 - \frac{r_s}{r} \right) dt^2$
...where $r_s$ says the distance from the center of the Earth, $r$ is the Earth's gravitational radius (if the Earth would be a non-rotating black hole, it would be its radius, it is 8.9mm). $d\tau$ is the proper time, i.e. the time what the clock measures, while $dt$ is the coordinate time (difference).
Substituting the values, we get:
$d\tau = \sqrt{1-\frac{8.9*10^{-3}}{6.3*10^6}}dt = \sqrt{1-1.4127*10^{-9}}dt = (1-7.063*10^{-10})dt$.
Thus, the time goes here with $1-7.063*10^{-10}$ "speed", compared to the deep space.
It is 1s delay in every 45 years.

Note: it is the gravitational time dilation, i.e. how is the time slowed due to gravity. The time delay due to speed (with the well-known $\frac{1}{\sqrt(1-\frac{v^2}{c^2})}$ formula is a different effect, and it depends on the speed.)
On the surface of the neutron stars, the time can be even 30-50% slower as far from them.
A: $$t'=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$$
You can calculate it yourself. Find out what high speed typically is at Earth - (choose the $v$ you wish to compare) - and plug it into the formula. Insert fx $t=100$ and the resulting $t'$ is the real actual time passed. It will be larger for larger speed. 
As an example, the ISS astronauts save $0.007$ s during half a year in orbit. 
