In Binary Tree a node can have at most $2$ children.
Total number of node $\mathbf{N} =$ node with $0$ child $+$ node with $1$ child $+$ node with $2$ child.
$\implies N= n_0 + n_1 + n_2 ($here, in question it is given that no. of leaf nodes i.e no. of nodes with $0$ children is $n$ so $n_0 = n)$
$\implies N=n + n_1 + n_2$
Total number of edges $e=N-1,$ and also $e = n \ast 0+ n_1 \ast1 + n_2 \ast 2$
$\therefore N-1 = n \ast 0+ n_1 \ast1 + n_2 \ast 2$
$\implies n + n_1 + n_2 - 1 = n_1 \ast1 + n_2 \ast 2$
$\implies \mathbf{n_2 = n -1}$
Option B is answer.
NOTE - For the tree, the degree of a node is defined as the number of sub-trees of the node or no of children of a node.