Let's analyze the problem step-by-step:
1. Let the side length of the large square countertop be $S$.
2. The smaller square tile inlay in the center has side length $s$, creating a uniform strip of width $w$ all around:

$$
\(w=\frac{S-s}{2}\)
$$

3. The area of the large square is:

$$
\(A_{\text {large }}=S^2\)
$$

4. The area of the smaller square tile is:

$$
\(A_{\text {tile }}=s^2\)
$$

5. The untiled strip area is:

$$
\(A_{\text {strip }}=A_{\text {large }}-A_{\text {tile }}=S^2-s^2\)
$$


Given the ratio of the tiled area to the untiled area is $25: 39$, we have:

$$
\(\frac{A_{\text {tile }}}{A_{\text {strip }}}=\frac{25}{39}\)
$$


Substitute the areas:

$$
\(\frac{s^2}{S^2-s^2}=\frac{25}{39}\)
$$

Rearranging:

$$
\(39 s^2=25\left(S^2-s^2\right) \Longrightarrow 39 s^2=25 S^2-25 s^2 \Longrightarrow 39 s^2+25 s^2=25 S^2 \Longrightarrow 64 s^2=25 S\)
$$


Now, find the strip width $w$ :

$$
\(w=\frac{S-s}{2}=\frac{S-\frac{5}{8} S}{2}=\frac{\frac{3}{8} S}{2}=\frac{3}{16} S\)
$$


The strip width is $\(\frac{3}{16}\)$ times the side length $S$.
So, the strip width is proportional to the side length of the countertop.
For the given options for the width of the strip-1.5, 3 , and 4.5 inches-the strip width must be $\(\frac{3}{16} S\)$.

Possible widths depend on $S$; if $S$ is such that $\(\frac{3}{16} S\)$ equals $1.5,3$, or 4.5 , then these could be the strip widths.

Check each:
- $\(w=1.5: S=\frac{16}{3} \times 1.5=\frac{16}{3} \times \frac{3}{2}=8\)$ inches
- $\(w=3: S=\frac{16}{3} \times 3=16\)$ inches
- $\(w=4.5: S=\frac{16}{3} \times 4.5=24\)$ inches

Since all calculated $S$ values are positive and realistic side lengths for a countertop, all values 1.5, 3 , and 4.5 inches could be the strip width.

Therefore, the correct answers are:
A) 1.5
B) 3
C) 4.5