To give an answer for the sake of it being here.
The Ricci tensor is defined as usual, as the trace of the Riemann curvature tensor.
Some authors use Ricci curvature to denote the function on the unit tangent bundle:
$$ UTM \ni v \mapsto \mathrm{Ric}(v,v)\in \mathbb{R} $$
The values of this function is sometimes called the Ricci curvatures.
Constant Ricci curvature at a point $p$ means that for all $v\in UT_pM$ we have $\mathrm{Ric}(v,v) = c$ for some constant $c$. In three dimensions because the Weyl curvature vanishes identically the Riemann curvature tensor is uniquely determined by the Ricci tensor, and the Ricci tensor being symmetric is uniquely determined by the Ricci curvature function. Hence you can in principle write the Riemann curvature tensor in terms of the Ricci curvature (by polarisation) and show that the sectional curvatures are constant.
Or you can use that in $\mathbb{R}^3$ the space of two-dimensional subspaces is itself three dimensional, so one can actually solve a linear system of equations to write sectional curvatures in terms of Ricci curvatures.
An important class of problems in Riemannian geometry is to understand the interaction between the curvature and topology.
An example of such interaction is given by the Gauss-Bonnet theorem which relates the geodesic curvature, the Gaussian curvature to the Euler characteristic of a regular surface of class $C^3$.
When studying the geometry of a smooth manifold we need to introduce the commutator of twice covariant differentiating vector fields which is called Riemannian curvature tensor.
As a matter of fact, if in Euclidean space we can change the order of differentiation, on a Riemannian manifold, the Riemann curvature tensor is in general nonzero.
For surfaces, the Riemann curvature tensor is equivalent to the Gaussian curvature K, which is scalar function.
On the other hand in dimensions larger than two, the Riemann curvature tensor is a tensor–field.
Now there are several curvatures associated to the Riemann curvature tensor.
Given a point $p \in M^n$ and two dimensional plane $\Pi$ in the tangent space of M at p, we can define a surface S in M to be the union of all geodesics passing through p and tangent to $\Pi$.
In a neighborhood of p, S is a smooth 2D submanifold of M. Then it is possible to define the sectional curvature $K(\Pi)$ of the 2D plane to be the Gaussian curvature of S at p.
Thus the sectional curvature $K$ of a Riemannian manifold associates to each 2D plane is a tangent space a real number.
You can imagine the sectional curvature as a kind of generalization of the Gaussian curvature.
Best Answer
You basically just have to look at the definitions: Given a unit-length tangent vector $x\in T_pM$, we obtain the Ricci curvature of $x$ at $p$ by extending $x=z_n$ to an orthonormal basis $z_1, \ldots, z_n$. And then $$\mathrm{Ric}_p(x) = \frac{1}{n-1}\sum_{i=1}^{n-1} \langle R(x,z_i)x,z_i\rangle$$ where $R$ denotes the Riemannian curvature tensor. On the other hand, the sectional curvature $K(x, z_i)$ for $i<n$ is given by (remember the $z_i$ are orthonormal) $$K(x, z_i) = \frac{\langle R(x,z_i)x,z_i\rangle}{\Vert x\Vert\, \Vert z_i\Vert-|\langle x, z_i\rangle|^2}=\langle R(x,z_i)x,z_i\rangle$$ Since $K(x,z_i) \ge C$ by your assumption, we obtain $$\mathrm{Ric}_p(x) = \frac{1}{n-1}\sum_{i=1}^{n-1} \langle R(x,z_i)x,z_i\rangle \ge C > 0$$ So you get a lower bound on the Ricci curvature.
Attention: All sign conventions are as can be found in DoCarmo's "Riemannian Geometry".