Two springs have their force constants k₁ and k₂ and they are stretched to the same extension. If k₂ > k₁ work done is

Due to the equal forces in both springs:

$\mathrm{F}={\mathrm{k}}_1{\mathrm{x}}_1={\mathrm{k}}_2{\mathrm{x}}_2$

From the question,

$k_1>k_2$

then, $\frac{\mathrm{F}}{{\mathrm{k}}_1}<\frac{\mathrm{F}}{{\mathrm{k}}_2}\Rightarrow {\mathrm{x}}_1<{\mathrm{x}}_2$

work = $\frac{1}{2}k{x}^2$

$\Rightarrow {\mathrm{W}}_1=\frac{{\mathrm{F}}^2}{2{\mathrm{k}}_1} \mathrm{and} {\mathrm{W}}_2=\frac{{\mathrm{F}}^2}{2{\mathrm{k}}_2}$

Thus, the second spring's work done is more than the first.

Hence the correct answer is  more in spring $k_2.$