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

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

$F_2=2.3 \times 10^5 \,N$