The required height of a TV tower which can cover the population of $$6.03$$ lakh is $$h$$. If the average population density is $$100$$ per square km and the radius of earth is $$6400 \text{ km}$$, then the value of $$h$$ will be ______ m.

The population covered is $$6.03 \text{ lakh} = 6.03 \times 10^5$$, the population density is $$100 \text{ per km}^2$$, and the radius of the Earth is $$R = 6400 \text{ km}$$.

From the population and density, the area covered by the TV tower is $$\text{Area} = \dfrac{\text{Population}}{\text{Population density}} = \dfrac{6.03 \times 10^5}{100} = 6030 \text{ km}^2$$.

For a TV tower of height $$h$$, the coverage area can be expressed as $$A = \pi d^2$$ where $$d = \sqrt{2Rh}$$; hence $$A = \pi \times 2Rh = 2\pi Rh$$.

Setting this equal to the area gives $$6030 = 2\pi \times 6400 \times h$$, so $$h = \dfrac{6030}{2\pi \times 6400} = \dfrac{6030}{40212.4} = 0.14998 \text{ km}$$, which is approximately $$h \approx 0.15 \text{ km} = 150 \text{ m}$$.

Therefore, the height of the TV tower is $$\boxed{150}$$ m.