= Solution
{c}
Expand the fourth power and take <expected values>. <independence> factors products involving distinct indices, and centering makes every term with an index occurring just once have <expected value> zero. The only surviving index patterns are four equal indices and two pairs. There are six orderings of each pair pattern, so
$$
\mathbb ES_n^4
=\sum_{i=1}^n\mathbb EX_i^4
+6\sum_{i<j}\mathbb EX_i^2\,\mathbb EX_j^2.
$$
Uniform boundedness gives $\mathbb EX_i^4\leq c^4$ and $\mathbb EX_i^2\leq c^2$. Hence
$$
\mathbb ES_n^4\leq nc^4+6\binom n2c^4
=(3n^2-2n)c^4\leq\boxed{3c^4n^2.}
$$
Identical distributions were never used: only <independence>, the common bound, and zero means are needed.
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