The electric potential (V) at a distance (r) from a point charge (q) is given by:

$$V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$$

where $$\frac{1}{4 \pi \varepsilon_0}$$ is Coulomb's constant.

In this case, we have:

• $$q = 4 \times 10^{-7} \mathrm{C}$$
• $$r = 9 \mathrm{~cm} = 0.09 \mathrm{~m}$$
• $$\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$$

Substituting these values into the equation for electric potential, we get:

$$V = (9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}) \frac{(4 \times 10^{-7} \mathrm{C})}{(0.09 \mathrm{~m})}$$

$$V = 4 \times 10^4 \mathrm{~V}$$

Therefore, the value of electric potential at a distance of $9 \mathrm{~cm}$ from the point charge $$4 \times 10^{-7} \mathrm{C}$$ is $$4 \times 10^4 \mathrm{~V}$$.

The correct answer is Option D.