Answer should be (B). As according to question, truth table will be like: $$\begin{array}{c c c c |c} \hline A & B & C & D & f \\\hline 0 & 0 & 0 & 0 & 0 \\ 0&0&0&1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & \text{don’t care} \\ 1 & 0 & 1 & 1 & \text{don’t care} \\ 1 & 1 & 0 & 0 & \text{don’t care} \\ 1 & 1 & 0 & 1 & \text{don’t care} \\ 1 & 1 & 1 & 0 & \text{don’t care} \\ 1 & 1 &1 & 1 & \text{don’t care} \end{array}$$
Using this truth table we get $3$ sub cube which are combined with following minterms $A (8,9,10,11,12,13,14,15)$, $BD( 5,13,7,15)$ and $BC(6,7,14,15)$
So, $f = A+ BD +BC= A+ B(C+D)$
So, minimum gate required $2$ OR gate and $1$ AND gate $= 3$ minimum gates.