Number of $’N’$ bit Mux/Demux/Decoder/Encoder to construct $’M’$ bit Mux/Demux/Decoder/Encoder is $ = \left \lceil \dfrac{M-1}{N-1} \right \rceil.$
Here, $M = 16,N = 4$
Then, number of decoders $ = \left \lceil \dfrac{16-1}{4-1} \right \rceil = \left \lceil \dfrac{15}{3} \right \rceil = 5.$
So, the correct answer is $5.$