1. We are given three resistors: $A=2\,\Omega,\quad B=4\,\Omega,\quad C=6\,\Omega$ We need the combination whose equivalent resistance is $\frac{22}{3}\,\Omega$

2. Check each option one by one.

Option A

Parallel combination of $A$ and $C$, connected in series with $B$.

First, equivalent of $A \parallel C$: $R_{AC}=\frac{A\cdot C}{A+C}=\frac{2\cdot 6}{2+6}=\frac{12}{8}=\frac{3}{2}\,\Omega$

Now add $B=4\,\Omega$ in series: $R_{eq}=\frac{3}{2}+4=\frac{3}{2}+\frac{8}{2}=\frac{11}{2}\,\Omega$

$\frac{11}{2}=5.5\,\Omega \neq \frac{22}{3}\,\Omega$ So, Option A is incorrect.

Option B

Parallel combination of $A$ and $B$, connected in series with $C$.

First, equivalent of $A \parallel B$: $R_{AB}=\frac{A\cdot B}{A+B}=\frac{2\cdot 4}{2+4}=\frac{8}{6}=\frac{4}{3}\,\Omega$

Now add $C=6\,\Omega$ in series: $R_{eq}=\frac{4}{3}+6=\frac{4}{3}+\frac{18}{3}=\frac{22}{3}\,\Omega$

This matches the required value. So, Option B is correct.

Option C

Series combination of $A$ and $C$, connected in parallel with $B$.

First, equivalent of $A$ and $C$ in series: $R_{AC}=2+6=8\,\Omega$

Now this is in parallel with $B=4\,\Omega$: $R_{eq}=\frac{8\cdot 4}{8+4}=\frac{32}{12}=\frac{8}{3}\,\Omega$

$\frac{8}{3}\,\Omega \neq \frac{22}{3}\,\Omega$ So, Option C is incorrect.

Option D

Series combination of $B$ and $C$, connected in parallel with $A$.

First, equivalent of $B$ and $C$ in series: $R_{BC}=4+6=10\,\Omega$

Now this is in parallel with $A=2\,\Omega$: $R_{eq}=\frac{10\cdot 2}{10+2}=\frac{20}{12}=\frac{5}{3}\,\Omega$

$\frac{5}{3}\,\Omega \neq \frac{22}{3}\,\Omega$ So, Option D is incorrect.

3. Therefore, the only suitable combination is: $\boxed{\text{Option B}}$

4. Comparison with stored correct answer:

• Stored correct answer: B
• Derived answer: B

So, the derived answer agrees with the stored correct answer.