The ammonia evolved from the treatment of $$0.30$$ g of an organic compound for the estimation of nitrogen was passed in $$100$$ mL of $$0.1$$ M sulphuric acid. The excess of acid required $$20$$ mL of $$0.5$$ M sodium hydroxide solution for complete neutralization. The organic compound is

The total amount of acid taken is calculated first.

For $$H_2SO_4$$:

$$\text{Molarity} = 0.1\text{ M}$$

Since $$H_2SO_4$$ is dibasic,

$$\text{Normality} = 0.1\times2=0.2\text{ N}$$

Therefore, the total milli-equivalents of acid taken are

$$\text{meq of }H_2SO_4=0.2\times100=20\text{ meq}$$

The unreacted acid is neutralized by $$NaOH$$.

For $$NaOH$$:

$$\text{Molarity} = 0.5\text{ M}$$

Since $$NaOH$$ is monobasic,

$$\text{Normality} = 0.5\times1=0.5\text{ N}$$

Hence,

$$\text{meq of }NaOH=0.5\times20=10\text{ meq}$$

Therefore, the acid that reacted with ammonia is

$$20-10=10\text{ meq}$$

Hence, the milli-equivalents of $$NH_3$$ evolved are also $$10\text{ meq}$$.

Using Kjeldahl's formula,

$$%\text{ Nitrogen}=\frac{1.4\times(\text{meq of }NH_3)}{\text{Mass of organic compound}}$$

Substituting the given values,

$$%\text{ Nitrogen}=\frac{1.4\times10}{0.30}$$

$$%\text{ Nitrogen}=\frac{14}{0.30}=46.66%$$

Now, calculate the percentage of nitrogen in each compound.

For acetamide, $$CH_3CONH_2$$:

$$\text{Molar mass}=59\text{ g mol}^{-1}$$

$$%\text{N}=\frac{14}{59}\times100\approx23.7%$$

For thiourea, $$NH_2CSNH_2$$:

$$\text{Molar mass}=76\text{ g mol}^{-1}$$

$$%\text{N}=\frac{28}{76}\times100\approx36.8%$$

For urea, $$NH_2CONH_2$$:

$$\text{Molar mass}=60\text{ g mol}^{-1}$$

$$%\text{N}=\frac{28}{60}\times100=46.66%$$

For benzamide, $$C_6H_5CONH_2$$:

$$\text{Molar mass}=121\text{ g mol}^{-1}$$

$$%\text{N}=\frac{14}{121}\times100\approx11.5%$$

The calculated nitrogen percentage matches that of urea.

Hence, the organic compound is $$\boxed{\text{Urea}}$$.

Therefore, the correct answer is Option (C).