An important property that helps in keeping the $B / B^+$ trees balanced is that they enforce at least 50% space utilization.
B-tree nodes are kept between 50 and 100 percent full
-- Navathe 6th Edition, page 647, line number 3.
This means if a node can accommodate max $p$ children, it must accommodate at least $\frac{p}{2}$ children at any time. (This is not applicable to the root, however.)
Of course, fractions are resolved by taking ceil, because if the $B / B^+$ tree wants us to have minimum $3.6$ children, we will keep $4$ children and not $3$. So, $\frac{p}{2}$ is actually $\left \lceil \frac{p}{2} \right \rceil$.
Now, we know that the number of keys is always one less than the number of children.
So minimum number of keys = $\left \lceil \frac{p}{2} \right \rceil-1$.
Now, coming to the question.
Max keys = 5. Hence, max children it can have = 6. (p is 6)
Minimum keys it can have = $\left \lceil \frac{p}{2} \right \rceil-1$.
= $\left \lceil \frac{6}{2} \right \rceil-1$.
= $2$
Option B