the correct answer is option C) \(F\over\sqrt{mk}\) 
Detailed solution:: 
Maximum speed is at mean position (equilibrium). F = kx
\(x={F\over k}\) 
\(W_F + W_{sp} = \Delta KE
\) 
\(F(x) – {1\over2} kx^2={1\over2}mv^2-0\) 
\(F\left({F\over k}\right) – {1\over2} k\left({F\over k}\right)^2={1\over2}mv^2\) 
\(\implies v_{max} = {F\over\sqrt{mk}}\)