Two metallic wires of identical dimensions are connected is series....

Let length of wire be $l$,
Area of wire as $A$
For equivalent wire length $= 2l$ & area will be $A$

As the wires are connected in series combination, the thermal resistance

${R}_{eq} = {R}_{1} + {R}_{2}$

$⇒ \dfrac{2l}{{\sigma}_{eq}A} = \dfrac{l}{{\sigma}_{1}A} + \dfrac{l}{{\sigma}_{1}A}$

$⇒ \dfrac{2}{{\sigma}_{eq}} = \dfrac{1}{{\sigma}_{1}} + \dfrac{1}{{\sigma}_{2}} ⇒ {\sigma}_{eq} = \dfrac{2{\sigma}_{1}{\sigma}_{2}}{{\sigma}_{1} + {\sigma}_{2}}$