How precise are BQP measurements?

Let's say I am given a Hamiltonian $$H$$, whose ground state is efficiently preparable. We know that $$||H|| \leq 1$$. Let that ground state be $$|\psi_{0}\rangle$$, with energy $$E_{0}$$. We also know that the unitary $$U$$ such that $$\begin{equation} U |\psi_{0}\rangle |00\cdots0\rangle = |\psi_{0}\rangle |E_{0}\rangle, \end{equation}$$ is efficiently preparable.

Consider the promise problem. For the yes instances, $$\begin{equation} E_{0} \geq c, \end{equation}$$ and for the no instances, $$\begin{equation} E_{0} \leq s, \end{equation}$$ where $$c - s \geq \frac{1}{O\big(2^{poly(n)}\big)}$$. Would this promise problem be in $$BQP$$?

As you've written it, yes. The implication of what you've written is that $$|E_0\rangle$$ is prepared exactly. Since it has been done efficiently, we know that you must have used only a number of qubits that's polynomial in $$n$$. Thus, I can measure those poly($$n$$) qubits and exactly determine the value of $$E_0$$. Once I have an exact classical value, it's no problem to compare it to $$c$$ and $$s$$, again each expressed using poly($$n$$) bits.