Consider a curve $\alpha:[a,b]\rightarrow\mathbb{R}^3$ where $T(t)=\frac{\alpha(t)}{v(t)}$ is the unitary tangent vector and $v(t)=|\alpha'(t)|$ is the scalar velocity.
The curvature of $\alpha$ is defined as $k(t)=\frac{T'(t)}{v(t)}$ and the radius of curvature as $\frac{1}{k(t)}$.
What is the geometric meaning of the curvature and how is the radius the inverse of the curvature?
The idea is to define our intuition of curvature mathematically, or, roughly speaking, how 'bent' that curve is. Of course, intuitively, the 'bendness' don't need to be uniform, i.e. a given curve $\alpha$ may have points which are 'more bent' then others. This is why the curvature is itself a function of $t$. Also, this is a nice definition since it is invariant under reparametrizations (of course, you do not expect the curvature to depend on how you parametrize your curve).
The idea becomes easier to understand if we take a circle as an example. If $\alpha(t) = (R\cos(t),R\sin(t))$, $R>0$ and $t\in [0,2\pi]$, you can check that the curvature $\kappa(t)$ is given by $$\kappa(t) = \frac{1}{R}$$
First, note that it is independent of $t$, which is intuitively right since the curvature of the circle does not chance as you change $t$. Also, it is the inverse of the radius. Why? Well, take two circles, with radius $R_{1}>R_{2}$. If you compare both circles (see a figure here) the circle with $R_{2}$ is indeed more curved than the other one.