Two springs of force constant 3000 N/m and 6000 N/m are stretched by same force. The ratio of their respective potential energies  is

$$\begin{gathered} Potential Energy PE of simple harmonic oscillator is \\ PE=\frac{1}{2}{\text{kx}}^2\text{, K=Force constant, x=displacement} \\ we know that restoring force of a spring is F=-kx, here k is force cons\tan t, x is displacement \\ \Rightarrow \text{ x=-F/k, Substitute this in PE,} \\ PE\text{=}\frac{\text{1}}{\text{2}}\text{k}{\left(\text{-}\frac{\text{F}}{\text{k}}\right)}^2 \\ \Rightarrow PE\text{=}\frac{\text{1}}{\text{2}}\text{k}\frac{{\text{F}}^{\text{2}}}{{\text{k}}^{\text{2}}} \\ \Rightarrow PE\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{F}}^{\text{2}}}{\text{k}} \\ \Rightarrow PE\alpha \frac{\text{1}}{\text{k}}, by question Force is same, hence taken cons\tan t, substitute force cons\tan t k_1,k_2 values \\ \Rightarrow \frac{P{E}_{\text{1}}}{P{E}_{\text{2}}}\text{=}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}=\frac{6000}{3000}=\frac{2}{1} \end{gathered}$$