To find the surface charge density of a charged metal cube, we need to determine the charge per unit area. The surface charge density, denoted by $\sigma$, is given by:

$\sigma = \frac{Q}{A}$

where:

• $ Q $ is the total charge on the cube
• $ A $ is the total surface area of the cube

The cube has a side length of $ 5 \mathrm{~cm} $, so each side of the cube is $ 5 \mathrm{~cm} $. Since 1 cm = 0.01 m, the side length in meters is:

$$5 \mathrm{~cm} = 0.05 \mathrm{~m}$$

The cube has 6 faces, and the area of one face is given by:

$$\text{Area of one face} = \left( 0.05 \mathrm{~m} \right)^2 = 0.0025 \mathrm{~m}^2$$

Therefore, the total surface area of the cube is:

$$A = 6 \times 0.0025 \mathrm{~m}^2 = 0.015 \mathrm{~m}^2$$

The total charge, $ Q $, is given as $ 6 \mu \mathrm{C} $. Converting this to Coulombs:

$$6 \mu \mathrm{C} = 6 \times 10^{-6} \mathrm{~C}$$

Now, plugging the values into the formula for surface charge density:

$$\sigma = \frac{6 \times 10^{-6} \mathrm{~C}}{0.015 \mathrm{~m}^2} = 4 \times 10^{-4} \mathrm{~C} \mathrm{m}^{-2}$$

Therefore, the surface charge density on the cube is:

$$\sigma = 0.4 \times 10^{-3} \mathrm{~C} \mathrm{m}^{-2}$$

So, the correct answer is:

Option D: $$0.4 \times 10^{-3} \mathrm{C} \mathrm{m}^{-2}$$