A 2 W carbon resistor is color coded with green, black, red and silver respectively. The maximum current which can be passed through this resistor is:

First we interpret the colour bands of the carbon resistor. The standard colour-digit table is:

$$\text{Black}=0,\; \text{Brown}=1,\; \text{Red}=2,\; \text{Orange}=3,\; \text{Yellow}=4,\; \text{Green}=5,\; \text{Blue}=6,\; \text{Violet}=7,\; \text{Grey}=8,\; \text{White}=9.$$

In the given sequence green, black, red, silver we have:

Green gives the first digit $$a = 5,$$

Black gives the second digit $$b = 0,$$

Red gives the multiplier power $$c = 2,$$ because red corresponds to the multiplier $$10^2$$,

Silver gives the tolerance of $$\pm 10\%.$$(We shall use this tolerance a little later.)

The standard formula for the resistance of a 4-band resistor is

$$R = (10a + b)\times 10^{\,c}\;\Omega.$$

Substituting the digits just obtained we get

$$R = (10\times 5 + 0)\times 10^{\,2}\;\Omega.$$

Simplifying step by step, first evaluate the bracket:

$$10\times 5 + 0 = 50.$$

Now multiply by the power of ten supplied by the third band:

$$R = 50 \times 10^{\,2}\;\Omega.$$

Since $$10^{\,2}=100,$$ we have

$$R = 50 \times 100\;\Omega = 5000\;\Omega.$$

So the nominal resistance is

$$R_{\text{nom}} = 5\;\text{k}\Omega.$$

Because the tolerance is silver (±10 %), the actual resistance can be as low as

$$R_{\min}= R_{\text{nom}}\left(1-0.10\right).$$

Substituting the numerical value,

$$R_{\min}= 5\,000\;\Omega \times 0.90 = 4\,500\;\Omega.$$

To ensure the resistor never dissipates more than its rated power of 2 W, we use the power relation

$$P = I^{2}R.$$

For the maximum current we must employ the minimum resistance (because $$I^{2}R$$ must stay at 2 W). Rearranging the formula for current gives

$$I_{\max} = \sqrt{\dfrac{P}{R_{\min}}}.$$

Now substitute $$P = 2\;\text{W}$$ and $$R_{\min}=4\,500\;\Omega$$:

$$I_{\max} = \sqrt{\dfrac{2}{4\,500}}\;\text{A}.$$

Compute the fraction first:

$$\dfrac{2}{4\,500}=0.000444\overline{4}.$$

Taking the square root,

$$I_{\max}= \sqrt{0.000444\overline{4}}\;\text{A}\approx 0.0211\;\text{A}.$$

Convert amperes to milliamperes using $$1\;\text{A}=1000\;\text{mA}$$:

$$0.0211\;\text{A}\times 1000 = 21.1\;\text{mA}.$$

The nearest value among the given options is 20 mA.

Hence, the correct answer is Option C.