Is black holes 'size' measure of its gravity? How can we talk about black holes size if it's a sizeless point of density striving for infinity, and measured is only the radius of optical manifestation of it's extremely hight gravity? 
Can I assume that 'size' of black hole is measure of its gravity?
 A: The singularity at the centre of a black hole is indeed a point, the Riemann tensor diverges there.
When we talk about the size of a black hole, we usually mean the event horizon, which is a function of the mass. The expression is:
$$r = \frac{2GM}{c^2},$$
where $G$ is the gravitational constant, $M$ is the mass of the black hole, and $c$ is the speed of light. Think of this as the point of no return, where the escape velocity of the black hole's gravitational field exceeds the speed of light.
A: The gravity of a black hole is the measure of by which a frame is transported. This is measured by the symmetries of the spacetime, which is with the Killing vectors. The killing vector for the Schwarzshild metric is $K_t~=~(1~-~2m/r)^{1/2}\partial_t$, and $m~=~GM/c^2$. The horizon radius is of course $r_s~=~2m$. The surface gravity is then
$$
\nabla^\mu(K_\nu K^\nu)~=~-2gK^\mu,
$$
which with the application of $\nabla_\mu$ it is not hard to see that the gravity $g$ is
$$
g^2~=~-\frac{1}{2}\nabla^\mu K^\nu \nabla_\mu K_\nu.
$$
For the Schwarzschild metric this 
$$
g~=~\frac{1}{1~-~2m/r}\frac{m}{r^2}.
$$
Clearly a tiny black hole can have a huge gravity close to the horizon. 
