$K$ is a field, $K^n$ is a vector space with $(e_1, \ldots, e_n)$. The tensor product $K^n \otimes K^n$ has the basis $\mathcal {B} = (e_1 \otimes e_1, \ldots, e_1 \otimes e_n, e_2 \otimes e_1,, \ldots, e_2 \otimes e_n, e_3 \otimes e_1, \ldots, e_n \otimes e_n)\,.$
Look at the matrices $A,B \in M(n \times n; K)$. We write $A \otimes B : K^n \otimes K^n \to K^n$. How does the block matrix look?
a. $A \otimes B = \begin {pmatrix} b_{11} A & b_{12} A & \cdots & b_{1n} A \\ b_{21} A & b_{22} A & \cdots & b_{2n} A \\ \vdots & \vdots &\ddots & \vdots \\ b_{n1} A & b_{n2} A & \cdots & b_{nn} A \end {pmatrix}$
b. $A \otimes B = \begin {pmatrix} a_{11} B & a_{12} B & \cdots & a_{1n} B \\ a_{21} B & a_{22} B & \cdots & a_{2n} B \\ \vdots & \vdots &\ddots & \vdots \\ a_{n1} B& a_{n2} B & \cdots & a_{nn} B \end {pmatrix}$
c. $A \otimes B = \begin {pmatrix} A & \cdots & A & 0 & \cdots & 0 \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ A & \cdots & A & 0 & \cdots & 0 \\ 0 & \cdots & 0 & B & \cdots & B \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & B & \cdots & B \end {pmatrix}$
Actually, the matrix form is just a representation of an abstract object. Let $C=A \otimes B $. It is obvious that C is a tensor with 4 indexes. $$C_{ijkl}=A_{ij} B_{kl}$$
If you take a look at your cases you will see that the correct answer is b).
By the way, a) represents $B \otimes A$.