1. In paper chromatography, the retardation factor $R_f$ is defined as:

$R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent front}}$

2. For substance A:

$R_f(A) = \frac{2.08}{3.25}$

3. For substance B:

$R_f(B) = \frac{1.05}{3.25}$

4. Therefore, the ratio of $R_f$ values is:

$\frac{R_f(A)}{R_f(B)} = \frac{\frac{2.08}{3.25}}{\frac{1.05}{3.25}} = \frac{2.08}{1.05}$

5. Now calculate:

$\frac{2.08}{1.05} \approx 1.98 \approx 2$

6. Hence, the required integer answer is:

$\boxed{2}$