Naive question about polynomial time reducibility 
Question:
  is it known, whether there is an upper bound on the exponent of the fastest polynomial-time reduction from $\mathrm{3SAT}$ to $\mathrm{NP}$-$\mathrm{hard}$ problems or, can it be proved that for every problem requiring an $\Theta(n^k)$ time reduction, there is a problem requiring an  $\Omega(n^{k+1})$ reduction from $\mathrm{3SAT}$?  

Additional, secondary question:
Which $\mathrm{NP}$-$\mathrm{hard}$ problem requires the polynomial time reduction from $\mathrm{3SAT}$ with the highest exponent?
 A: It's unlikely there is any upper bound.  To see the problem, consider the (artificial) problem
$\text{3SATpad} = \{ \phi \#^{|\phi|^{100}} : \phi \in \text{3SAT}\}$, 
where $\#$ is some new symbol.  $\text{3SATpad}$ is in $\mathsf{NP}$, and there is an obvious $O(n^{100})$-time reduction from $\text{3SAT}$ to $\text{3SATpad}$.  But it's doubtful there's even an $O(n^{99})$-time reduction.  Otherwise, one could solve $\text{3SAT}$ in $2^{O(n^{.99})}$ time (unlikely, as this violates the Exponential Time Hypothesis), as follows:


*

*On input $\phi$ of length $n$, run the reduction producing string $y$.  The key point here is that $|y| \leq O(n^{99})$.

*If $y$ is not of the form $\psi \#^{|\psi|^{100}}$, reject.  

*Otherwise, since $|\psi \#^{|\psi|^{100}}| \leq O(n^{99})$, it must be that $|\psi| \leq O(n^{.99})$.  Now decide if $\psi \in \text{3SAT}$ in $2^{O(n^{.99})}$ time, using a naive brute-force algorithm.  This gives the correct answer about satisfiability of $\phi$.
Perhaps one can even prove the impossibility conditioned only on $\mathsf{P} \neq \mathsf{NP}$ (rather than on E.T.H.), by a Ladner's Theorem-type argument...
