Any noncomputable real trivially satisfies your question. Here's an explicit construction of a *computable* real which is hard to compute:

Let $\{\varphi_e\}$ list all (partial) computable functions, and $\{p_e\}$ all polynomials.

We define a real number $r$ as follows: the $n$th bit of the binary expansion of $r$ is a $1$ iff $\varphi_i(n)$ does **not** halt and output $1$ in $\le p_j(n)$ steps (so, either doesn't halt in that time, or does halt and outputs something $\not=1$) - where $n=\langle i, j\rangle$. (Here "$\langle\cdot,\cdot\rangle$" denotes the [Cantor pairing function](https://en.wikipedia.org/wiki/Pairing_function).)

$r$ is computable (note that we run each computable function for only a bounded amount of time before deciding what to do for that bit), but $r$ is not computable in polynomial time since it diagonalizes against all polynomial-time computations: if $\varphi_i$ has running time bounded by $p_j$, then $\varphi_i(\langle i, j\rangle)\not=r(\langle i, j\rangle)$.

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Note that nothing special about polynomials was used here: given any reasonable complexity class (for instance, any complexity class of the form "Running time bounded by some $f\in\mathcal{F}$" for $\mathcal{F}$ a computable set of computable total functions), there is a computable real whose bits are not computable by any algorithm in that class.