Enumerate all squares with $1, \ldots, 90$. Let the number of rice grains promised for the $n$-th square be $F_n$; then according to the problem $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all $n > 2$. Notice that $F_n > F_{n-1}$ if $n > 2$, hence $F_{2(n+1)} = F_{2n+2} = F_{2n+1} + F_{2n} > F_{2n} + F_{2n} = 2 \cdot F_{2n}$ for all $n$. This implies $F_{2 \cdot 4} > 2 \cdot F_{2 \cdot 3} = 2 \cdot 8 = 2^4$ and by mathematical induction $F_{2n} > 2^n$ for all $n > 3$. Therefore $F_{90} > 2^{45} > 2^{44} = (2^{11})^4 = 2048^4 > 2015^4$.