L={anbn∣n≥0}L={anbn∣n≥0}
The given Language is Context free language and CFL are closed under Union, Concatenation, and kleen closure,
S1 : $L^2$ is a context free language, here $L^2$ can be writtern as
$L^2$ = L.L and CSL is closed under concatenation so $L^2$ is a CFL
S2:$L^k$ context free language for any given k≥1, $L^k$ can be writtern as L.L.L…..k times and CFL are closed under concatenation so $L^k$ is CFL
S3: L¯ and $L^∗$ are context free languages
Context free language are not closed under complementation for complement of L may be CFL or CSL.
here complement of L is CFL,
and CFL are closed under kleen closure (*) so it is CFL.
so correct ans is D.