= Solution
Let $G_0=\mathbb F_p^n$ and let
$$
\Gamma=\{(x,\phi(x)):x\in G_0\}\subseteq G_0\times G_0.
$$
For each fixed first coordinate $d$, the second coordinates occurring in $\Gamma-\Gamma$ are values of $\phi(x+d)-\phi(x)$, of which there are at most $C$. Therefore
$$
|\Gamma-\Gamma|\leq C|G_0|=C|\Gamma|.
$$
The difference-set form of the <Plünnecke-Ruzsa inequality> now gives
$$
|3\Gamma-2\Gamma|\leq C^5|\Gamma|.
$$
Every $u\in X$ has the form
$$
(0,u)=(x,\phi(x))-(x+a,\phi(x+a))-(x+b,\phi(x+b))+(x+a+b,\phi(x+a+b)),
$$
so $\{0\}\times X\subseteq2\Gamma-2\Gamma$. Consequently
$$
(\{0\}\times X)+\Gamma\subseteq3\Gamma-2\Gamma.
$$
The set on the left has exactly $|X||\Gamma|$ elements, since its fiber over each $x$ is $\phi(x)+X$. It follows that
$$
|X||\Gamma|\leq C^5|\Gamma|,
$$
and hence $|X|\leq C^5$.
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