Does it make sense to have $\operatorname{Ricci}(g)=3$? Is there a context in which it would make sense to have $\operatorname{Ricci}(g)$ to equal a constant number, such as $3$?
I am not sure this makes sense, but apparently it does.
 A: If by "$\operatorname{Ricci}(g)$" you mean the usual Ricci tensor (which is a symmetric $2$-tensor field), then no, $\operatorname{Ricci}(g)=3$ doesn't make sense.
However, there is a way to define a "Ricci function"; let's call it $\rho_g$. Given a Riemannian manifold $(M,g)$, let $SM\subset TM$ be the unit tangent bundle (the set of all tangent vectors of unit length), and define $\rho_g\colon SM\to\mathbb R$ by
$$
\rho_g(v) = \operatorname{Ricci}(g)(v,v).
$$
Given a unit vector $v\in T_pM$, $\rho_g(v)$ is the sum of sectional curvatures of all planes spanned by $(v,E_i)$ as $E_i$ ranges over an orthonormal basis for the subspace of $T_pM$ orthogonal to $v$.  (See my Riemannian Manifolds, p. 147.) Thus if $M$ has constant sectional curvature $C$, then $\rho_g$ is the constant function $(n-1)C$ (where $n=\dim M$).
It's a straightforward exercise using the polarization identity to show that $\rho_g \equiv 3$ if and only if $\operatorname{Ricci}(g) = 3g$.  This is what people mean when they make an informal statement like "$\operatorname{Ricci}(g) = 3$."
A: If ${\rm Ric}\ (g)=3g$ then $$ {\rm Ric}\ (kg)=3g $$ for any $k>0 $ so that $$ {\rm Ric}\ (h)=h $$ where $h=3g$
