Depends on your eye. You can realise the curvature of the Earth by just going to the beach. Last summer I was on a scientific cruise in the Mediterranean. I took two pictures of a distant boat, within an interval of a few seconds: one from the lowest deck of the ship (left image), the other one from our highest observation platform (about 16 m higher; picture on the right):
The distance d$d$ from an observer O$O$ at an elevation h$h$ to the visible horizon follows the equation (adopting a spherical Earth):
d=Ratan(sqrt(2R*h)/R)$$
d=R\times\arctan\left(\frac{\sqrt{2\times{R}\times{h}}}{R}\right)
$$
where d$d$ and h$h$ are in meters and R=6370e3 m$R=6370*10^3m$ is the radius of the Earth. The plot is like this:
But addressing more precisely the question. Realising that the horizon is lower than normal (lower than the perpendicular to gravity) means realising the angle (gamma$gamma$) that the horizon lowers below the flat horizon (angle between OH$OH$ and the tangent to the circle at O, see cartoon below; this is equivalent to gamma in that cartoon). This angle depends on the altitude h$h$ of the observer, following the equation:
gamma=180/PIatan(sqrt(2R*h)/R)$$
\gamma=\frac{180}{\pi}\times\arctan\left(\frac{\sqrt{2\times{R}\times{h}}}{R}\right)
$$
where gamma$\gamma\ is in degrees, see the cartoon below.
