Allow me to add some engineer hand-waving.
If the gravitational dilation factor ##f = t_0/t_\infty## (i.e. local clock rate relative to clock rate at infinite) at distance ##r## from the center of a spherical gravitational field of mass ##m## can be found as $$f = \sqrt{1-\frac{2 G m}{r c^2}}$$ then for ##f## near 1 we can, to ease the numerical work, use the approximation $$f \approx 1 - \frac{Gm}{r c^2} = 1 - k, $$ where we now can work with ##k## values instead.
Further, we can get the dilation factor for the surface of a planet around a star relative to infinity, for instance for Earth and Jupiter in orbit around the Sun, by multiplying the factor for position in the planets field with the factor for the position in the stars field.
If I punch those ##k##-numbers into a spreadsheet then for Earth I get ##k_E \approx 1.1\cdot10^{-8}## (combined from a surface and orbit ##k## around ##6.9\cdot10^{-10}## and ##9.8\cdot 10^{-9}##) and for Jupiter I get ##k_J \approx 2.3\cdot 10^{-8}## (combined from a surface and orbit ##k## around ##2.1\cdot 10^{-8}## and ##1.9\cdot 10^{-9}##).
Dividing the dilation factors for Earth and Jupiter (both relative to infinity) means that the relative clock rate at Jupiter surface relative to Earth surface should be ##t_J/t_E \approx 1-1.2\cdot 10^{-8}.## That means a clock on Jupiter surface would loose around 1 sec every 930 days relative to a clock on Earth due to gravitational dilation alone.
