The required number is $c = 4/3$. We first show that if $n$ is a positive integer and $x_1, \dots, x_n$ are real numbers greater than or equal to $-1$ such that $\sum_{k=1}^{n} x_k^3 = 0$, then $\sum_{k=1}^{n} x_k^2 \le 4n/3$. Indeed, since $x_k^3 - 3x_k^2 + 4 = (x_k + 1)(x_k - 2)^2 \ge 0$, $k = 1, \dots, n$, it follows that $\sum_{k=1}^{n} x_k^2 \le 4n/3 + (1/3) \sum_{k=1}^{n} x_k^3 = 4n/3$.

The latter inequality is strict, unless $n$ is divisible by $9$, in which case equality holds if and only if $8n/9$ numbers are $-1$ and the remaining $n/9$ are all $2$. The conclusion follows.