Q:

Definition of event horizon – gravity around black holes

Put simpler:
A black hole is surrounded by a 'sphere' where, to an outside observer, the speed of light is (near) zero.
What is the radius of that 'sphere' (for a non-rotating black hole) and how strong is gravity there?
Definition of Event Horizon (EH)
A black hole is defined as a body whose mass falls within its Schwarzschild
radius $r_\mathrm S$, where the escape velocity $v_\mathrm e$ equals the speed of light $c$.
That is clear, nothing can escape from it. But, with the Event Horizon (EH)
described as a boundary that cannot be crossed, EH does not coincide with $r_\mathrm S$.
At $r_\mathrm S$, and everywhere in a gravitational field, only the acceleration of
gravity $g$ is apparent. $v_\mathrm e$ is derived, the equivalence of endless
decelerations by $g$. A particle does not directly experience $v_\mathrm e$.
Some arithmetic with $v_\mathrm e=(2GM/r)^{1/2}$ and $g=GM/r^2$, shows, in round numbers:

A 'small' black hole of 3.3 solar masses ($M_\mathrm s$) – $r_\mathrm S$ is $10\ \mathrm{km}$ and at $r_\mathrm S$,
$g$ is a crushing $4.5\times10^9\ \mathrm{km/s^2}$. Its sphere of no return, where
$g$ is $300\,000\ \mathrm{km/s^2}$ is further out at $1200\ \mathrm{km}$.
For one comparable to our Milky Way's of $3.3\times10^6M_\mathrm s$, $r_\mathrm S=1.0\times10^7\ \mathrm{km}$ with
$g$ at $r_\mathrm S$ a puny $4500\ \mathrm{km/s^2}$. Its EH is closer in at $1.2\times10^6\ \mathrm{km}$.
The only instance where $g$ equals $300\,000\ \mathrm{km/s^2}$ at $r_\mathrm S$ is for a black hole
of $50\,000M_\mathrm s$ with $r_\mathrm S$ at $1.5\times10^5\ \mathrm{km}$.

Clearly I'm missing something.

A:

It is a truly unfortunate coincidence that the "escape velocity" at the Schwarzschild radius $R_\mathrm{S}$ is $c$. This leads people to think Newtonian mechanics can predict black holes. It can't.
Note that a Newtonian escape velocity is the minimum velocity needed to coast to infinity against gravity. If you have less speed, you still travel away from the surface some amount before being pulled back down, whereas nothing is able to move any distance at all out of the event horizon. Also, one can always escape a Newtonian gravity source at any speed with a rocket, whereas no rocket can help escape the event horizon.
So what is the event horizon? It is the surface that separates parts of spacetime that can transmit a signal to infinity from those that cannot. Technically one has to know the details of all the gravity and matter and energy in all of spacetime to know the location of the event horizon (you might think you can transmit a signal, but what if you are inside a black hole and don't yet realize it?). We only have simple formulas in very simply hypothetical cases (e.g. a single black hole in an empty universe that has been sitting unchanging for all eternity).
Acceleration doesn't have much to do with event horizons. You will need to accelerate more to hover above a small black hole compared to a large one, but that's about it. Note that $c = 300{,}000\ \mathrm{km/s}$, but an acceleration of $300{,}000\ \mathrm{km/s^2}$ has no special significance, just as a distance of $300{,}000\ \mathrm{km}$ is not a special distance.

