There were a number of partial solutions but this well explained one is almost entirely the work ofAndrei of Tudor Vianu National College and shows how "obvious" the answer is with a little visualising and careful reasoning.

In the first solution, I have 4 cylinders (the 4 squares length $l-4r$and $2r$) and 8 half-cylinders - cylinders cut through the diagonal (the 8 small right-angled triangles from the figure).

Each of the four cylinders has the volume: $V_1=\pi r^2(l-4r)$

Each of the half-cylinders has a volume of half a cylinder: $V_2=\frac{1}{2}\times \pi r^2\times 2r=\pi r^3$ Now, the total volume of the tubular stand is: $V=4\times V_1+8\times V_2=4\pi r^2(l-2r)$

With the second method, I arrange the 8 half-cylinders to make 4 bigger cylinders, and these 4 to the bigger ones. The height of one 'big' cylinder is $(l-2r)$, and the total volume is: $V=4\pi r^2(l-2r)$

which is exactly the result obtained above. Substituting the numerical values: $l=10\mathrm{cm}$and $r=.5\mathrm{cm}$, I obtain:

$V=4\pi \times 0.25\times (10-1)=9\pi =28.27{\mathrm{cm}}^3$

If the volume wood would be double $(18\pi )$, then the outside dimension of the dowel would be:

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{18\pi }& =& 4\pi \times 0.25(l-1)\\ \multicolumn{1}{c}{l-1}& =& 18\mathrm{cm}\\ \multicolumn{1}{c}{l}& =& 19\mathrm{cm}\end{array}}$

If the volume of wood would be the same but the radius would be 1 cm, then the outside dimension would be:

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{9\pi }& =& 4\pi (l-2)\\ \multicolumn{1}{c}{l-2}& =& \frac{9}{4}\\ \multicolumn{1}{c}{l}& =& 4.25\mathrm{cm}\end{array}}$

The general formula for the volume of the dowel (proved above) is:

$V=4\pi r^2(l-2r)$

One could see that the volume is proportional to the square of the radius, and in the limit of long outside dimensions, proportional to the outside dimension.