To solve this problem, we need to understand the relationship between the current, the drift velocity, and the cross-sectional area of the wire. The electric current $I$ in a wire is given by

$I = n e A v_d$

where:

• $n$ is the number density of electrons,
• $e$ is the charge of an electron,
• $A$ is the cross-sectional area of the wire, and
• $v_d$ is the mean drift velocity of the electrons.

Let's denote the current in the thicker wire as $$I_1 = 100 \mathrm{~mA} = 0.1 \mathrm{~A}$$ and the current in the thinner wire as $$I_2 = 200 \mathrm{~mA} = 0.2 \mathrm{~A}$$.

The diameter of the thicker wire is $d$, so its cross-sectional area, $A_1$, is

$$A_1 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$$

For the thinner wire, the diameter is $\frac{d}{2}$, so its cross-sectional area, $A_2$, is

$$A_2 = \pi \left(\frac{d}{4}\right)^2 = \frac{\pi d^2}{16}$$

Using the relation for current in the thicker wire, we have

$$I_1 = n e A_1 v = n e \left(\frac{\pi d^2}{4}\right) v = 0.1 \mathrm{~A}$$

Let the mean drift velocity in the thinner wire be denoted by $v'$. For the thinner wire:

$$I_2 = n e A_2 v' = n e \left(\frac{\pi d^2}{16}\right) v' = 0.2 \mathrm{~A}$$

Now, we can equate the expressions for current and solve for $v'$:

$$\frac{\pi d^2}{16} n e v' = 0.2 \mathrm{~A}$$

$$v' = \frac{0.2 \mathrm{~A}}{n e \left(\frac{\pi d^2}{16}\right)}$$

From the equation for the thicker wire:

$$0.1 \mathrm{~A} = \frac{\pi d^2}{4} n e v$$

$$v = \frac{0.1 \mathrm{~A}}{n e \left(\frac{\pi d^2}{4}\right)}$$

Dividing the two equations, we get

$$\frac{v'}{v} = \frac{0.2 \mathrm{~A}}{\left(\frac{\pi d^2}{16}\right) n e} \times \frac{(\frac{\pi d^2}{4}) n e}{0.1 \mathrm{~A}} = \frac{0.2 \mathrm{~A} \times 4}{0.1 \mathrm{~A} \times 16} = \frac{4}{1} = 8$$

Thus,

$v' = 8 v$

The mean drift velocity of electrons in the thinner wire is 8 times the mean drift velocity in the thicker wire. Therefore, the correct option is:

Option B: $8 v$