Why does electrical resistivity have units of $\Omega \cdot \mathrm{m}$ rather than $\Omega \cdot \mathrm{m}^3 ?$ Electrical resistivity has units of $\Omega \cdot \mathrm{m} .$  However, since resistivity can be described as the resistance of a unit cube, shouldn't the units therefore be $\Omega \cdot \mathrm{m}^3$ instead?
I ask after seeing this question to which the answer is apparently $\left(\text{D}\right) :$

$
\text{Resistivity can be described correctly as:} \\
\hspace{1em}
\begin{array}{cl}
\mathbf{A} & \text{resistance of a unit length.} \\
\mathbf{B} & \text{resistance per unit area.}   \\
\mathbf{C} & \text{resistance per unit volume.} \\
\mathbf{D} & \text{resistance of a unit cube.}
\end{array}
$

 A: The resistance $R$ of a body grows with bigger length $l$
(a longer wire has greater resistance)
and shrinks with bigger cross-section area $A$ (a thicker wire has smaller resistance).
Hence you have
$$ R = \rho\frac{l}{A}$$
and resistivity $\rho$ must have unit $\Omega\cdot$m.
A: Resitivity can be thought of as resistance of a unit cube, but for a unit cube, the resistance doesn't work out to $\text{material constant} * \text{volume}$.
Instead, restivity ($\rho$) is given by $\rho = \frac {RA}{L}$ (where $R$ is resistance, $A$ is area and $L$ is length of material) or to rearrange in terms of net resistance $R = \frac {\rho L}{A}$.
We can see that this suggests that resistance will increase with length, but decrease with area.  This should make sense, because if you send the same current through a wider area, it should experience less resistance, and if you push the same current through a longer path of resistance, it should experience more net resistance over the path.
You can see from that relationship that the units will work out to resistance per unit length, not per unit volume.
