Logs (but not the mathematical kind)

Since the log is uniform, we can immediately ignore the height of the log and focus on its cross section only.

Let $x$ represent the radius of the non-bark component of the log. This yields the following diagram.

The area of the non-bark component of the log is given by $π\times x^{2}$ , while the area of the area of the total log is given by $π\times 10^{2}=100π$ . We can combine these to a single equation: $\frac{π\times x^{2}}{100π}=\frac{1}{4}$ . (Note: we get $\frac{1}{4}$ from the area of the whole log - the area of the bark: $1 - \frac{3}{4}=\frac{1}{4}$ ).

$π$ is present in both the numerator and denminator so cancels, and further simplification yields $4x^{2}=100$ .

$x^{2}=25$

$x=\sqrt{25}=5$ (we take the positive root only since the question concerns length).

This shows that the radius of the non-bark component is $5$ . We can then find the width of the bark by minusing this from the total log radius, yielding $10-5=5$ . So the solution is $\boxed{5}$ .

Alternatively, you could solve the problem with scale factors. $ASF=LSF^{2}$ , and $\sqrt{\frac{1}{4}}=\frac{1}{2}$ . $10-\frac{1}{2}\times 10=5$ .