We know

\(\frac{\text { Force } \times \text { Length }}{\text { Area } \times \text { young's modulus }}=\) elongation \(\quad\left\{\frac{F L}{A y}=\Delta x\right\}\)

\(\Rightarrow F=\left(\frac{\Delta x \cdot y}{L}\right) A\)

\(F=\left(\frac{\Delta x \cdot y}{L} \cdot \frac{\pi}{4}\right) d^2\)

We can say \(F \propto d^2\)

So we can use

\(\frac{F_1}{F_2}=\frac{d_1^2}{d_2^2}\)  \(\left\{\begin{array}{l}F_1=4 \times 10^5 N \\ d_1=2 mm \\ F_2=? \\ d_2=1.5 mm \end{array}\right\}\)

Substituting values

\(\frac{4 \times 10^5}{F_2}=\frac{(2)^2}{(1.5)^2}\)

\(F_2=2.3 \times 10^5 \,N\)