Answer is $16$. Proof sketch: Let $\ell$ be distance from $A$ to $B$. Let $d_i$ be (number of) vertices at distance $i$ from $A$. Then each shortest path from $A$ to $B$ takes one vertice from $d_i$. Therefore the total number of shorted paths is at most $d_1d_2 \cdots d_{\ell - 1}$. $$d_1d_2 \cdots d_{\ell - 1} \geq 100.$$Now it is classical that this implies $$d_1 + d_2 + \cdots d_{\ell - 1} \geq 13.$$The only case of equality is when $\{d_1,...,d_{\ell - 1}\} = \{2,2,3,3,3\}$ or $\{3,3,3,4\}$. However, in both cases you cannot have all edges between $d_i$ and $d_{i + 1}$ for all $0 \leq i \leq \ell - 1$ (assume $d_{0} = \{A\},d_{\ell} = \{B\}$), otherwise you would form $108$ shortest paths. Therefore you must remove some edge, at which point you can form at most $99$ shortest path. Thus you must have $$d_1 + d_2 + \cdots d_{\ell - 1} > 13.$$Construction: connect these graphs: $K_{1,4}$ + ($K_{4,3}$ - any 2 edge) + $K_{3,3}$ + ($K_{3,4}$ - any 2 edge) + $K_{4,1}$.