In the density measurement of a cube, the mass and edge length are measured as $$(10.00 \pm 0.10)$$ kg and $$(0.10 \pm 0.01)$$ m, respectively. The error in the measurement of density is:

We have a cube whose mass is given as $$m = 10.00 \ \text{kg} \pm 0.10 \ \text{kg}$$ and whose edge length is $$l = 0.10 \ \text{m} \pm 0.01 \ \text{m}$$. We wish to find the percentage (or fractional) error in its density.

First, recall the formula connecting mass, volume and density. Density is defined by

$$\rho = \frac{m}{V}.$$

Because the body is a cube, its volume is the cube of its edge length, so

$$V = l^3.$$

Substituting this expression for volume in the density definition gives

$$\rho = \frac{m}{l^3}.$$

When we need the error in a quantity which is a product or quotient of measured quantities raised to powers, we employ the law of propagation of relative (or fractional) errors:

If $$Q = a^p b^q$$, then the relative error in $$Q$$ is

$$\frac{\Delta Q}{Q} = |p|\frac{\Delta a}{a} + |q|\frac{\Delta b}{b}.$$

In our present case $$\rho = m \, l^{-3}$$, so the exponents are $$+1$$ for $$m$$ and $$-3$$ for $$l$$. Only their magnitudes matter. Thus the relative error in density is

$$\frac{\Delta \rho}{\rho} = 1 \times \frac{\Delta m}{m} + 3 \times \frac{\Delta l}{l}.$$

We now compute each individual relative error:

For mass,

$$\frac{\Delta m}{m} = \frac{0.10}{10.00} = 0.010.$$

For length,

$$\frac{\Delta l}{l} = \frac{0.01}{0.10} = 0.10.$$

Substituting these values into the propagation formula, we get

$$\frac{\Delta \rho}{\rho} = 0.010 + 3 \times 0.10.$$

Multiplying out the second term,

$$0.010 + 0.30 = 0.310.$$

Therefore the fractional (or percentage, if multiplied by 100) error in the measurement of density is

$$\frac{\Delta \rho}{\rho} = 0.31.$$

Since the options list numerical values of the error without units, the answer corresponds to $$0.31$$.

Hence, the correct answer is Option A.