A telephonic communication service is working at a carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz?

We have a carrier frequency of $$f_c = 10 \text{ GHz}$$. Converting gigahertz to hertz, we multiply by $$10^9$$, so $$f_c = 10 \times 10^9 \text{ Hz}$$.

Only 10% of this carrier frequency is actually used as the transmission bandwidth. Writing 10% as the decimal $$0.10$$, the available bandwidth becomes

$$B = 0.10 \times f_c = 0.10 \times 10 \times 10^9 \text{ Hz} = 1 \times 10^9 \text{ Hz}.$$

Thus the service can exploit a total bandwidth of $$1 \text{ GHz}$$ (since $$1 \text{ GHz} = 10^9 \text{ Hz}$$).

Each telephonic channel needs $$5 \text{ kHz}$$ of bandwidth. Using $$1 \text{ kHz} = 10^3 \text{ Hz}$$, this requirement is

$$\text{Bandwidth per channel} = 5 \times 10^3 \text{ Hz}.$$

The number of channels that can fit into the total bandwidth is found by simple division:

$$N = \frac{B}{\text{Bandwidth per channel}}.$$

Substituting the values, we get

$$N = \frac{1 \times 10^9}{5 \times 10^3}.$$

First we handle the powers of ten: $$\frac{10^9}{10^3} = 10^{9-3} = 10^6.$$ The expression simplifies to

$$N = \frac{1}{5} \times 10^6.$$

Now, $$\frac{1}{5} = 0.2$$, and $$0.2 \times 10^6 = 2 \times 10^5.$$

Therefore, the total number of telephonic channels that can be transmitted simultaneously is

$$N = 2 \times 10^5.$$

Hence, the correct answer is Option D.