These tree like structures are called the Dendogram which is useful for the similarity measurement in the cluster analysis as well as to detect the outliers.
The similarity between two objects in a Dendogram is represented as a height of the lowest internal node they share.
In a Single linkage (nearest neighbor) method, the distance between the two clusters is determined by the distance of the two closest objects (nearest neighbors) in the different clusters.
Now, comes to the given problem.
Here, we are given $5$ data points $x_1,x_2,x_3,x_4,x_5$.
Now, the least distance between these data points is $1$ that is between $x_1$ and $x_2$ $\&$ $x_4$ and $x_5$
It means $x_1$ and $x_2$ $\&$ $x_4$ and $x_5$ will form $2$ clusters. First cluster is having data points $x_1$ and $x_2$ $\&$ second cluster is having data points $x_4$ and $x_5.$
So, either $(A)$ is correct or $(B)$ is correct.
Now, due to single linkage, we have to find the distance of $x_3$ between the above two clusters.
(1) Distance between $x_3$ and $[x_1,x_2]:$
$d(x_3,[x_1,x_2])=\min(d(x_1,x_3),d(x_2,x_3))=\min(4,3)=3$
(2) Distance between $x_3$ and $[x_4,x_5]:$
$d(x_3,[x_4,x_5])=\min(d(x_4,x_3),d(x_5,x_3))=\min(2,5)=2$
So, $x_3$ will be merged with cluster $[x_4,x_5]$
Therefore,
Answer is A.