Let $M \in \{0,1\}^{n \times n}$ and let $r_i$ be its $i$-th row. Given constant $p \in (0,1/2]$, let the number of $1$s in each row be at least $p\,n$. Given constant $c \in (0,1)$, what is the maximum number $R_{max}$ of rows $r$ such that 

$$r\, r_i^{\top} \ge c\,n$$ 

holds for less than $n^{1/3}$ row indices $i$? Do we have
$R_{max}=\Theta(n^{1/3})$ for any constant $c>0$ and $p \in (0,1/2]$ when $n$ approaches infinity? Thank you!

---
Think about having the first $n^{1/3}−1$ rows with all of their $1$s in the first $pn$ columns, the second $n^{1/3}−1$ rows with all their $1$s in the second $pn$ columns, and so on. This construction allows me to see that, for any constant $c>0$ and $p \in (0,1/2]$, we have 
$$R_{max} \ge (1/p−1)(n^{1/3}−1)$$
It is not clear to me whether there exists a matrix construction showing that 
$$R_{max}=\omega\left((1/p−1)(n^{1/3}−1)\right)$$
for some constant $c>0$ when $n$ approaches infinity.

Perhaps, the most central question is: Do we have
$$R_{max}=\Theta\left((1/p-1)(n^{1/3}-1)\right)=\Theta(n^{1/3})$$
for any constant $c>0$ and $p \in (0,1/2]$ when $n$ approaches infinity?