Answer
$F=4.48\times10^4N$
Work Step by Step
$F=-ky$
$k=\frac{F}{-y}=\frac{760N}{6\times10^-5m}=1.267\times10^6\frac{N}{m}$
$mgy_i=mgy_f+\frac{ky_f^2}{2}$
$y_f^2+\frac{2mgy_f}{k}-\frac{2mgy_i}{k}=0$
$y_f^2+\frac{2(760N)y_f}{1.267\times10^6\frac{N}{m}}-\frac{2(760N)(1.0m)}{1.267\times10^6\frac{N}{m}}=0$
$y_f^2+1.2\times10^{-3}y_f-1.2\times10^{-3}=0$
Solving the quadratic equation, we get the two solutions: $y_f=100.8m$ and $y_f=0.037m$
$F=ky_f=(1.267\times10^6\frac{N}{m})(0.037m)=4.48\times10^4N$