In case of the series combination

$$\begin{gathered} ∆l=l_1+l_2 \\ Y=\frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{∆l}{l}}} \\ ∆l=\frac{l}{Y} \end{gathered}$$

After joining, the rod's equivalent length is 2l

Given that lengths and forces are equal in a series,

$$\begin{gathered} ∆l=∆l_1+∆l_2 \\ \frac{l_{eq}}{Y_{eq}}=\frac{l}{Y_1}+\frac{l}{Y_2} \\ \frac{2l}{Y}=\frac{l}{Y_1}+\frac{l}{Y_2} \\ Y=\frac{2Y_1{Y}_2}{Y_1+Y_2} \end{gathered}$$

Hence the correct answer is $Y=\frac{2Y_1{Y}_2}{Y_1+Y_2}.$