Let $n=mk$ where $2 \leq m, k \in \mathbb{Z}$ and let $A=[a_{ij}]$ be $n×n$ matrix given by $a_{ij} =1$ if for some
$r=0,1,...m-1, \ rk < i, j \leq (r+1)k$
and $a_{ij}=0$ otherwise. Then the nullspace of $A$ has dimension
1) $m(k-1)$
2) $mk-1$
3) $k(m-1)$
4) $0$
I took $n= 3(m)×2(k)$
$r=0, 0 < i, j \leq 2$ gives $a_{11} , a_{12}, a_{21}, a_{22}=1$ and $ a_{13}, a_{14}= 0$;
$r=1, 1 < i , j \leq 4$ gives $a_{22} =a_{23}=a_{24}=a_{32}=a_{33}=a_{34}=1$ and $ a_{35}= a_{36}= a_{53}=a_{56}=0$;
$r=2, 2< i , j \leq 6$ gives $a_{22} , a_{35}=a_{36}=a_{45}=a_{46}=a_{64}=a_{65}=a_{66}=1$.
Therefore the matrix is $ A= \left[ {\begin{array}{cccccc} 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1 & 1 & 1\\ 0 & 0 & 1 & 1 & 1 & 1\\ 0 & 0 & 1 & 1 & 1 & 1\\ \end{array} } \right] $ whose rank is 4 and therefore $dim(N(A))=2$ which is given by option 3. But how to prove this in general?
Your matrix should look like this:$$A= \begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 1 & 1\\ \end{bmatrix}$$
Given $a_{ij}=1$ for $rk<i,j\le rk+k$. For each $r$, we have $k$ admissible values for both $i,j$, i.e. $\{rk+1,rk+2,...,rk+k\}$. These values represent a square sub-matrix of order $k$ between $a_{rk+1,rk+1}$ and $a_{rk+k,rk+k}$. Each value of $r$ gives you a distinct submatrix, no two of which share a row or column. The rank of each submatrix is $1$, and since we have $m$ such submatrices, the rank of $A$ is $m$. Thus,$$\dim\ker A=mk-m$$