1. Given: Two metallic wires are connected in series and have identical dimensions.

   Let each wire have:

   • length $l$
   • cross-sectional area $A$

   Their conductivities are $\sigma_1$ and $\sigma_2$.

2. Resistance of each wire

   We know: $R = \frac{l}{\sigma A}$

   So, $R_1 = \frac{l}{\sigma_1 A}, \qquad R_2 = \frac{l}{\sigma_2 A}$

3. Equivalent resistance in series

   Since the wires are in series: $R_{\text{eq}} = R_1 + R_2 = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A}$

   $R_{\text{eq}} = \frac{l}{A}\left(\frac{1}{\sigma_1} + \frac{1}{\sigma_2}\right)$

4. Dimensions of equivalent single wire

   Since two identical wires are joined in series:

   • total length $= 2l$
   • cross-sectional area remains $A$

   If effective conductivity is $\sigma_{\text{eq}}$, then $R_{\text{eq}} = \frac{2l}{\sigma_{\text{eq}} A}$

5. Equate the two expressions

   $\frac{2l}{\sigma_{\text{eq}} A} = \frac{l}{A}\left(\frac{1}{\sigma_1} + \frac{1}{\sigma_2}\right)$

   Cancel $\frac{l}{A}$ from both sides: $\frac{2}{\sigma_{\text{eq}}} = \frac{1}{\sigma_1} + \frac{1}{\sigma_2}$

   $\frac{2}{\sigma_{\text{eq}}} = \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2}$

   Therefore, $\sigma_{\text{eq}} = \frac{2\sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$

6. Compare with options

   This matches Option B.

Final Answer: $\boxed{\frac{2\sigma_1 \sigma_2}{\sigma_1 + \sigma_2}}$