Science, 2012 Apr 27, p441 speaks of clock accuracies of 10^{−18}.
The Schwarzschild metric for the Earth is

dτ^{2} = (1 − r_{s}/r)dt^{2} + …

dτ = (1 − r_{s}/r)^{½}dt ≃ (1 − r_{s}/(2r))dt + …

∂τ/∂t = (1 − r_{s}/(2r))
= (the time dilation due to gravity well),

r_{s} = 2GM_{⦿}/c^{2}
= 2 ∙ 6.7 ∙ 10^{−11} m^{3}kg^{−1}s^{−2}
∙ 6 ∙ 10^{24}kg / (3 ∙ 10^{8} m/s)^{2}
= 0.009 m ≃ 1 cm

r_{s} is about 1 cm for the mass of the Earth.

r = 7 ∙ 10^{3} km = 7 ∙ 10^{6} m
(Earth radius)

∂(∂τ/∂t)/∂r = ∂^{2}τ/∂t∂r
= ∂(1 − r_{s}/(2r))/∂r
= r_{s}/(2r^{2})

Near the Earth’s surface the radial gradient of the time dilation is ∂^{2}τ/∂t∂r = 0.01 m/(1.4 ∙ 10^{13} m^{2})
= 7 ∙ 10^{−16} m^{−1}

Thus when r is changed by 1cm = 10^{−2} m, (∂r = 10^{−2} m), the time dilation is changed by 7 ∙ 10^{−18} and such a clock, one cm above another, should see the difference.