The density of a cylindrical wire is given by the formula:

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

where $m$ is the mass, $r$ is the radius, and $l$ is the length.

The relative error in a calculated quantity is the sum of the relative errors in the quantities it depends on. For the density, this is given by:

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

Given that $\Delta m = 0.01$ g, $m = 0.4$ g, $\Delta r = 0.03$ mm, $r = 6$ mm, $\Delta l = 0.04$ cm, and $l = 8$ cm, we can substitute these values into the formula to find the relative error in the density:

$$\frac{\Delta \rho}{\rho} = \frac{0.01}{0.4} + 2\frac{0.03}{6} + \frac{0.04}{8} = 0.025 + 0.01 + 0.005 = 0.04$$

So the relative error in the density is 0.04, or 4%.