Tension is same (given)

From free body diagram

\(3 T=150 \,N\)

\(T=50 \,N\)

Since the bar has to be supported symmetrically Therefore extension in each wire will be same

We know \(\Delta x=\frac{F L}{A Y}\)

Compare \(1\) copper wire with another steel wire

\(\frac{F L}{A_C Y_C}=\frac{F L}{A_S Y_S}\)

\(\Rightarrow \frac{A_S}{A_C}=\frac{Y_C}{Y_S}\)  \(\left\{\begin{array}{l}\text { Where, } \\ A_C-\text { Area of copper wire } \\ Y_C-\text { Young's modulus copper } \\ A_S-\text { Area of steel wire } \\ Y_S-\text { Young's modulus steel }\end{array}\right.\)

Substitutuing value of \(Y_C\) and \(Y_s\)

\(\frac{\pi d_S^2}{4 \times \frac{\pi}{4} \times d_C^2}=\frac{110 \times 10^9}{190 \times 10^9}\)

\(\frac{d_S}{d_C}=\sqrt{\frac{11}{19}}\)  \(\left\{\begin{array}{l}d_S \text {-diameter of steel wire } \\ d_C \text {-diameter of copper wire }\end{array}\right.\)

\(\frac{d_C}{d_S}=\sqrt{\frac{19}{11}}\)