= Solution
Let $W_i=U_i^2$. Applying the same <measurable function> separately to <independent and identically distributed> <random variables> preserves those two properties, and
$$
\mathbb E[W_i]=\int_0^1u^2\,\mathrm du=\frac13.
$$
In particular, these bounded <random variables> are <integrable random variables>. The <weak law of large numbers> gives
$$
\overline W_n=\frac1n\sum_{i=1}^nW_i\xrightarrow{\mathbb P}\frac13.
$$
If $c>1/3$, then $\{\overline W_n>c\}\subseteq\{|\overline W_n-1/3|\geqslant c-1/3\}$, whose <probability> tends to zero. If $0\leqslant c<1/3$, then $\{\overline W_n\leqslant c\}\subseteq\{|\overline W_n-1/3|\geqslant1/3-c\}$, whose <probability> also tends to zero. Taking a complement in the second case proves
$$
\boxed{\mathbb P\!\left(\sum_{i=1}^nU_i^2>cn\right)\longrightarrow\begin{cases}0,&c>1/3,\\1,&0\leqslant c<1/3.\end{cases}}
$$
The positive gap from $1/3$ is essential to this argument. <Convergence in probability> alone does not determine the answer at the excluded threshold $c=1/3$.
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