Please explain question no.23 in the attached image???

Dear Student,
The capacitance finally is Ceff.
Let the K finally is keff
Ceff=keffe₀A/D

For the above 2 dielectrics they are in series.
effective capacitance of the above two series connected dieelectrics  is
1/C1 = D/k1e₀A + D/k2e₀A
=>C1=e₀A(k1k2)/D(k1+k2)

C2= Capacitance of the below dieelectric
=2k3e₀A/D
 Ceff= C1+C2
=e₀A(k1k2/(k1+k2) + 2k3)/D
 keff=(k1k2/(k1+k2) + 2k3)  Answer is C.