One has an upper bound of $\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $, and this is best possible - i.e. one can obtain $\mathbb E(U)$ arbitrarily close to this.

To see this, let's first consider the case where we know the $X_i$ are integer-valued with mean $\mu$ but the variance is unrestricted. Then we can obtain $\mathbb E(U) =\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $ by a distribution with probability $1- (\mu - \lfloor \mu \rfloor)$ on $\lfloor \mu \rfloor $ and $(\mu - \lfloor \mu \rfloor)$ on $\lfloor \mu \rfloor +1$.

This is optimal since we are trying to maximize $\sum_{n=1}^\infty \mathbb P(X \geq n)^6$ given $\sum_{n=1}^\infty \mathbb P(X \geq n)=\mu$, for which increasing the larger values of $\mathbb P(X \geq n)$ and decreasing the smaller values always gives an improvement, so an optimum is obtained when the larger values are all $1$ and can't be increased and the smaller values are all $0$ and can't be decreased, i.e. when the probability distribution is supported on at most two values.

Let's now see that we can achieve $\mathbb E(U)$ arbitrarily close to $\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $ with a given variance $\nu$. Our previous construction works for $\nu= (\mu - \lfloor \mu \rfloor) (1- (\mu - \lfloor \mu \rfloor))$, and it is not possible to have variance smaller than this, so it suffices to handle the case when $\nu$ is larger.

If we shift $\epsilon$ probability mass from $\lfloor \mu \rfloor$ to $\lfloor \mu \rfloor+1$ and $\epsilon/m$ mass from $\lfloor \mu \rfloor$ to $\lfloor \mu \rfloor+m$, we have not changed the mean but have raised the variance by $\epsilon (m+1)$. Taking $$\epsilon = \frac{\nu- (\mu - \lfloor \mu \rfloor) (1- (\mu - \lfloor \mu \rfloor)) }{ m+1} $$

we see that for $m$ sufficiently large, the distribution is still well-defined after the mass shift, and taking $m$ sufficiently large we may take $\mathbb E(U)$ arbitrarily close to its initial value.