Yield strength of mild steel is given as
[tex]Y = 250 MPa[/tex]
here we know that
[tex]Y = \frac{F}{A}[/tex]
[tex]250 * 10^6 = \frac{F}{\pi r^2}[/tex]
now we can rearrange it as
[tex]F = \pi r^2 * 250* 10^6[/tex]
here we know that
[tex]r = 1.25 mm = 1.25 * 10^{-3} m[/tex]
[tex]F = \pi*(1.25*10^{-3})^2* 250* 10^6[/tex]
[tex]F = 1.23 * 10^3 N[/tex]
so maximum applied force will be F = 1.23 * 10^3 N
