To determine the maximum percentage error in the density of the wire, follow these steps:

The density of the wire is given by the formula for a cylinder:

$$\rho = \frac{m}{\pi r^2 l}$$

When calculating the maximum percentage (relative) error, you add the relative errors from each variable, taking into account the power to which each variable is raised. Thus, for density:

$$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$$

Here:

$$\frac{\Delta m}{m}$$ is the relative error in the mass.

$$2\frac{\Delta r}{r}$$ is due to the radius being squared.

$$\frac{\Delta l}{l}$$ is the relative error in the length.

Now plug in the given values:

Mass: $$m = 0.60 \, \text{g}$$ with an error $$\Delta m = 0.003 \, \text{g}$$

$$\frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005$$ or 0.5%.

Radius: $$r = 0.50 \, \text{cm}$$ with an error $$\Delta r = 0.01 \, \text{cm}$$

$$\frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02$$ or 2%. Since the radius is squared in the formula, multiply by 2:

$$2\frac{\Delta r}{r} = 2 \times 0.02 = 0.04$$ or 4%.

Length: $$l = 10.00 \, \text{cm}$$ with an error $$\Delta l = 0.05 \, \text{cm}$$

$$\frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005$$ or 0.5%.

Sum these contributions to find the overall maximum relative error:

$$\frac{\Delta \rho}{\rho} = 0.005 + 0.04 + 0.005 = 0.05$$

Converting this relative error into a percentage gives:

$$0.05 \times 100\% = 5\%$$

Thus, the maximum percentage error in the density of the wire is 5%, which corresponds to Option C.