The volume charge density of a sphere of radius $$6 \text{ m}$$ is $$2 \mu C \text{ cm}^{-3}$$. The number of lines of force per unit surface area coming out from the surface of the sphere is ______ $$\times 10^{10} \text{ N C}^{-1}$$.
[Given: Permittivity of vacuum $$\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$$]

Given: Radius of sphere $$R = 6 \text{ m}$$, volume charge density $$\rho = 2 \text{ } \mu\text{C cm}^{-3} = 2 \text{ C m}^{-3}$$.

(Converting: $$2 \text{ } \mu\text{C cm}^{-3} = 2 \times 10^{-6} \times 10^{6} \text{ C m}^{-3} = 2 \text{ C m}^{-3}$$)

The number of lines of force per unit surface area is the electric field at the surface of the sphere.

Using Gauss's law for a spherical surface:

$$E \times 4\pi R^2 = \dfrac{Q}{\varepsilon_0} = \dfrac{\rho \times \dfrac{4}{3}\pi R^3}{\varepsilon_0}$$

$$E = \dfrac{\rho R}{3\varepsilon_0}$$

Substituting the values:

$$E = \dfrac{2 \times 6}{3 \times 8.85 \times 10^{-12}} = \dfrac{12}{26.55 \times 10^{-12}}$$

$$E = 0.4520 \times 10^{12} = 4.52 \times 10^{11} \text{ N C}^{-1}$$

$$E = 45.2 \times 10^{10} \text{ N C}^{-1} \approx 45 \times 10^{10} \text{ N C}^{-1}$$

Therefore, the answer is $$\boxed{45}$$.