To determine the spring constant $$k$$ of a spring experimentally, we can use the formula derived from Hooke's Law and the period of oscillation for a mass-spring system:

$$T = 2 \pi \sqrt{\frac{m}{k}}$$

Here, $$T$$ is the period of oscillation, $$m$$ is the mass, and $$k$$ is the spring constant. Rearranging the formula to solve for $$k$$, we get:

$$k = \frac{4 \pi^2 m}{T^2}$$

To find the error in $$k$$, we have to consider the errors in both the measurements of $$T$$ and $$m$$. Let's denote the percentage errors as follows:

$$\Delta T / T \cdot 100\% = 2\%$$ (positive error)

$$\Delta m / m \cdot 100\% = -1\%$$ (negative error)

According to the rules of error propagation, the relative error in $$k$$ can be found by adding the relative errors in the measurements, each multiplied by the respective powers to which they affect $$k$$. Since $$T$$ is squared in the denominator and $$m$$ is linear in the numerator, the calculation is as follows:

$$\frac{\Delta k}{k} = \left| -2 \cdot \frac{\Delta T}{T} \right| + \left| 1 \cdot \frac{\Delta m}{m} \right|$$

Substituting the percentage errors:

$$\frac{\Delta k}{k} = \left| -2 \cdot 0.02 \right| + \left| 1 \cdot (-0.01) \right|$$

$$\frac{\Delta k}{k} = 0.04 + 0.01$$

$$\frac{\Delta k}{k} = 0.05$$

Thus, the percentage error in determining the value of $$k$$ is:

$$\frac{\Delta k}{k} \cdot 100\% = 5\%$$

The correct answer is Option A: 5%