, 2024

All six faces of the prism are painted which means that the $1$ by $1$ by $1$ cubes in the interior of the prism are
the only cubes that have no paint on them.

Each of the three dimensions of the prism (length, width, height) must
be at least $3$, otherwise there are
no $1$ by $1$ by $1$ cubes without paint on them.

The set of interior $1$ by $1$ by $1$ cubes must also be in the shape of a
rectangular prism.

(You should confirm each of these last two sentences for yourself before
reading on.)

There are $50$ interior $1$ by $1$ by $1$ cubes, and so the volume of the
interior prism is $50$.

Thus, we are looking for three positive integers, representing the
length, width and height of the interior prism, whose product is $50$.

We may use the positive divisors of $50$ ($1$, $2$, $5$, $10$, $25$, $50$) to help identify the four
possibilities: $1\times1\times50$,
$1\times2\times25$, $1\times5\times10$, and $2\times5\times5$.

These are the only ways to express $50$ as the product of three positive
integers.

Next, we determine the dimensions of the original prisms given each set
of dimensions for the interior prisms.

Consider the interior prism with dimensions $1\times1\times50$.

Recall that this is the prism that remains after all exterior (painted)
cubes are removed.

That is, $1$ by $1$ by $1$ cubes have been removed from the top
and bottom of the $1\times1\times50$
interior prism, from the left and right sides, as well as from the two
ends (the front and back).

This means that the dimensions of the original prism are each $2$ greater than the dimensions of the
interior prism. (You should try to visualize this.)

We complete the following table to determine the dimensions and the
volume of the original prism in each case.

Interior prism dimensions
Original prism dimensions
Volume of original prism

$1\times1\times50$
$3\times3\times52$
$V=3\times3\times52=468$

$1\times2\times25$
$3\times4\times27$
$V=3\times4\times27=324$

$1\times5\times10$
$3\times7\times12$
$V=3\times7\times12=252$

$2\times5\times5$
$4\times7\times7$
$V=4\times7\times7=196$

Therefore, the mean of all possible values of $V$ is $\dfrac{468+324+252+196}{4}=\frac{1240}{4}=310$.