1. We need an equivalent resistance of $R_{eq}=\frac{46}{3}\,\Omega$

2. The phrase in each option means:

• first two resistors are connected in parallel,
• and that parallel combination is in series with the remaining two resistors.

So for each option, $R_{eq}=R_{\text{parallel}}+R_{\text{series part}}$

3. Check option A:

Parallel part: $4\Omega$ and $6\Omega$ $R_p=\frac{4\cdot 6}{4+6}=\frac{24}{10}=\frac{12}{5}\,\Omega$

Series part: $2\Omega+8\Omega=10\Omega$

So, $R_{eq}=10+\frac{12}{5}=\frac{50+12}{5}=\frac{62}{5}\,\Omega$

This is not $\frac{46}{3}\Omega$.

4. Check option B:

Parallel part: $6\Omega$ and $8\Omega$ $R_p=\frac{6\cdot 8}{6+8}=\frac{48}{14}=\frac{24}{7}\,\Omega$

Series part: $2\Omega+4\Omega=6\Omega$

So, $R_{eq}=6+\frac{24}{7}=\frac{42+24}{7}=\frac{66}{7}\,\Omega$

This is not $\frac{46}{3}\Omega$.

5. Check option C:

Parallel part: $2\Omega$ and $6\Omega$ $R_p=\frac{2\cdot 6}{2+6}=\frac{12}{8}=\frac{3}{2}\,\Omega$

Series part: $4\Omega+8\Omega=12\Omega$

So, $R_{eq}=12+\frac{3}{2}=\frac{24+3}{2}=\frac{27}{2}\,\Omega$

This is not $\frac{46}{3}\Omega$.

6. Check option D:

Parallel part: $2\Omega$ and $4\Omega$ $R_p=\frac{2\cdot 4}{2+4}=\frac{8}{6}=\frac{4}{3}\,\Omega$

Series part: $6\Omega+8\Omega=14\Omega$

So, $R_{eq}=14+\frac{4}{3}=\frac{42+4}{3}=\frac{46}{3}\,\Omega$

This matches the required value.

7. Therefore, the correct option is $\boxed{D}$