Answer
The average value: $\bar f = \frac{1}{8}$.
Work Step by Step
We have $f\left( {x,y,z} \right) = xyz$ and ${\cal W}:0 \le z \le y \le x \le 1$.
The inequality suggests that the region ${\cal W}$ is bounded by the lines $z=y$, $y=x$ and $x=1$.
To find the description of ${\cal W}$ we sketch the corresponding solid region. Using the figure attached we can project ${\cal W}$ onto the $xy$-plane to obtain the domain ${\cal D}$. We see that ${\cal D}$ is bounded by the line $x=1$ and the line $y=x$. Notice that ${\cal D}$ is both vertically and horizontally simple region. We choose to describe ${\cal D}$ as a vertically simple region, so the description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le x} \right\}$
Thus, the region ${\cal W}$ consists of all points lying between ${\cal D}$ and the upper face, which is the plane $z=y$. Therefore, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 1,0 \le y \le x,0 \le z \le y} \right\}$
So, the volume of ${\cal W}$:
$Volume\left( {\cal W} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^x \mathop \smallint \limits_{z = 0}^y {\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^x \left( {z|_0^y} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^x y{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {{y^2}|_0^x} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 {x^2}{\rm{d}}x$
$ = \frac{1}{6}\left( {{x^3}} \right)|_0^1 = \frac{1}{6}$
To find the average value of $f$, we use Eq. (5):
$\bar f = \frac{1}{{Volume\left( {\cal W} \right)}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V$
$\bar f = 6\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^x \mathop \smallint \limits_{z = 0}^y xyz{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = 3\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^x xy\left( {{z^2}|_0^y} \right){\rm{d}}y{\rm{d}}x$
$ = 3\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^x x{y^3}{\rm{d}}y{\rm{d}}x$
$ = \frac{3}{4}\mathop \smallint \limits_{x = 0}^1 x\left( {{y^4}|_0^x} \right){\rm{d}}x$
$ = \frac{3}{4}\mathop \smallint \limits_{x = 0}^1 {x^5}{\rm{d}}x$
$ = \frac{3}{{24}}\left( {{x^6}|_0^1} \right) = \frac{1}{8}$
Thus, the average value: $\bar f = \frac{1}{8}$.
