First of all the $k(G)$ cannot be smaller than the size of any proper subgroup of $G$, because if $H$ is a proper subgroup, $gH$ is a coset, $g\not\in H$, then $gHgH$ does not contain intersect $gH$ (it will not contain if $g$). ghgh'=gh''$, then $g\in H$). For example, if $G$ is Abelian, $|G|$ is not prime (i.e. $G$ is not cyclic of prime order), then $k(G)>\sqrt{|G|}$: look at the size of a maximal proper subgroup.
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First of all the $k(G)$ cannot be smaller than the size of any proper subgroup of $G$, because if $H$ is a proper subgroup, $gH$ is a coset, $g\not\in H$, then $gHgH$ does not contain $gH$ (it will not contain $g$). For example, if $G$ is Abelian, $|G|$ is not prime (i.e. $G$ is not cyclic of prime order), then $k(G)>\sqrt{|G|}$: look at the size of a maximal proper subgroup. |
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