What is $\text{kg}\cdot \text{m}^2$ concretely? (multiplication of units) I have problem to understand what $\text{kg}\cdot \text{m}^2$ (moment of inertia) is. So, for example a force does a work of $3\text{J}=3\text{Nm}$ means that the force can displace a weight of $3\text{N}$ by $1\text{m}$ against gravity or equivalently will displace a weight of $1\text{N}$ by $3\text{m}$. A moment of $3\text{Nm}$ mean that to rotate the system I have to apply a force of $3\text{N}$ at $1\text{m}$ of the point of equilibrium or equivalently $1\text{N}$ at $3\text{m}$ of the equilibrium point.
What  does $3\text{ kg}\cdot \text{m}^2$ mean?
 A: The moment of inertia is given by$^*$
$$I=\int r^2\ \text d m$$
In other words, we are adding up the squared distance to some axis for each mass element, but the sum is weighted by the mass of each element $\text d m$. 
The unit tells us that $I$ depends on both mass and how that mass is distributed throughout the system. It also gives us somewhat of a "proportionality", showing that the distance the mass is from the defined axis has more of an influence on $I$ than the mass itself does (due to the squared length as opposed to the mass to the first power).
So for $3\ \rm{kg\cdot m^2}$ it's "the sum of square distance each mass is from the axis (length$^2$) weighted by the mass (mass) that totals to $3\ \rm{kg\cdot m^2}.$" Or, if you wanted to think of it in terms of a single particle, you could say something like "A particle with mass $1\ \rm{kg}$ that is a squared distance of $3\ \rm{m^2}$ from the axis in question.", but this isn't as general as the moment inertia usually is. In other words, $3\ \rm{kg\cdot m^2}$ could be true for many different systems.

$^*$This is actually a simplification of the complete moment of inertia tensor, which I will not go into here.
A: If you want a tangible interpretation like in your example with Nm you should think about how any two objects with the same moment of inertia compare to each other. In the end it's more of an abstract quantity since it is usally defined as an abbreviation in the treatment of rotation for a compound system of an arbitrary amount of particles (see Aaron Stevens' answer for the mathematical expression and explanation). To compare two different objects in 3d you actually need the full tensor, which further adds to the abstractness of this quantity.
Nevertheless you could view the moment of inertia as a meausre of how a specific total mass is distributed in a given circular area around the axis of rotation (since m$^2$ only depends on distance to the axis) if you compare objects of similar size. Or you can compare objects of similar mass and see how this mass distribution circle differs. It actually depends on what part of the mass is located on which ring of given distance $r$ around the axis but the total moment of inertia doesn't tell you anything about the shape of the object.
The full tensor yields an ellipsoid as a geometric relation, where each of the main axes correspond to one main moment of inertia. Therefore, what I've said above is not usefully applicable for anything that doesn't have a fixed axis of rotation, since different objects only have the same rotational properties if they have the same ellipsoid associated with them.
A: By the analogy to linear motion:
$$F = ma \Rightarrow m = \frac{F}{a}$$
with the units
$$1 \text{ kg} = \frac{1\text{ N}}{\frac{1 \text{ m}}{1 \text{ s}^2}}.$$
A body with mass $1 \text{ kg}$ requires $1 \text{ N}$ of force to be accelerated by $\frac{1 \text{ m}}{1 \text{ s}^2}$.
For rotations we have 
$$M = I \alpha \Rightarrow I = \frac{M}{\alpha}$$
with torque $M$, moment of inertia $I$ and angular acceleration $\alpha$. For the simplest case of a point particle and the force acting perpendicular to line that connects the mass and the center of the rotation, $M = Fr$ when $r$ is the distance between the particle and the center. The angular acceleration has the unit $\frac{1}{\text{s}^2}$, so for the unit of $I$ we get
$$[I] = \frac{[F][r]}{[\alpha]} = \underbrace{\frac{\text{kg} \cdot \text{m}}{\text{s}^2}}_{[F]} \cdot \underbrace{\text{m}}_{[r]} \cdot \underbrace{\text{s}^2}_{[1/\alpha]} = \text{kg}\cdot\text{m}^2.$$
One can say that the unit results from cancelling the seconds squared from the force and the angular acceleration.
A body with a moment of inertia of 3 $\text{kg}\cdot\text{m}^2$ requires


*

*$3\text{ N}$ of force $1\text{ m}$ away from the center to get an angular acceleration of $1 \text{ rad}$ per second squared or

*$1\text{ N}$ of force $3\text{ m}$ away from the center to get an angular acceleration of $1 \text{ rad}$ per second squared or

*$1\text{ N}$ of force $1\text{ m}$ away from the center to get an angular acceleration of $1/3 \text{ rad}$ per second squared or

*any other combination so that force times distance divided by angular acceleration is 3.

