Consider a point mass $A$ with mass $m$ in empty space. The point mass $A$ does not have a velocity and does not rotate. Since gravity is symmetric for nonmoving objects, the spacetime curvature around $A$ is also symmetric.
So at a distance $d$ from the point mass $A$ how strong is the curvature $C$ ?
$$ C = f(d,m) $$ $$f = ???$$
It sounds as if you just want the acceleration given by the non-relativistic equation from Newton's law:
$$ a = \frac{GM}{r^2} $$
where $M$ is the mass of the object generating the gravitational field (strictly speaking this equation only applies when the mass of the accelerating object is much less than $M$).
For the GR version of this have a look at twistor59's answer to What is the weight equation through general relativity?:
$$ a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} $$
This is the simplest treatment of the problem I've seen, but even so I suspect you'll have problems with this unless your maths is reasonably advanced.
