Is this problem based on a known hard problem?
This problem hugely depends on the distributions of the $c_{(i)}$, if they are independant or not, also on the size of the space in which each coordinate of the vectors are, and on the information you gain with each check if $c_{(i)}$ is correct.

Assuming the vector space is $Z_q^n$ and $n$ is polynomial in $\log q$, that $c_{(i)}$ are independant and uniformly distributed, and that your verification algorithm only tells you if $c_{(i)}$ is correct or not, let us look at the problem for one instance of $a$, $b$ and $c$. 

If you take $b$ uniformly at random, you have probability $\frac{1}{q}$ to have $⟨a,b⟩=c$, which is exponentially small. 

Moreover, there is no way to improve brute force on the value of $⟨a,b⟩$ because you don't learn anything beside the correctness of your $b$, so this is a problem you would call hard is cryptography. 
Even if you have a polynomial number of instances, this won't help you that much because finding a $b$ such that even only one $⟨a_i,b_i⟩=c_i$ is already exponentially hard.

Of course, this is an information theoretical approach to the complexity of this problem, and I think the real assumption you want to reduce to is one that would allow you to use my results, i.e. an assumption which makes you learn nothing when you verify $⟨a,b⟩=c$.

I know this is not really a reduction, but that's all I've got to help you, I hope it is enough.

Let me edit to reply to yours.

First, say you have $q$ groups of $a^{(i)}$ and $c^(i)$.

Now if you know $c$, you are just asking to solve a system of $qn$ linear equations with $n^2$ unknown. Now I think the subject is quite known and you can find a lot of documentation on it easily.

Remark that $B$ being an orthogonal matrix corresponds to adding $n$ equations to the system. Even though they are no more linear, I think the subject has already been studied.

As for the complexity questions, if I'm not wrong you can solve the problem with an algorithm running in time polynomial in $n$, so one would consider it easy.