Yes, this is totally feasible. Unless $p$ is purposely chosen as a safe prime, there's a good chance that's easy.
Computing $a$ given $B$, $p$, and $C=B^a\bmod p$ is a Discrete Logarithm Problem in the multiplicative group modulo prime¹ $p$, of order $r=p-1$.
For any DLP problem, there is a generic attack of cost a few times $\sqrt r$ group operations, Pollard's rho. Not only is $2^{\approx45}$ multiplications modulo 90-bit $p$ feasible, but that can be efficiently distributed on independent machines, see Paul C. van Oorschot and Michael J. Wiener's Parallel Collision Search with Cryptanalytic Applications (in Journal of Cryptology, 1999, following their article in proceedings of CCS 1994).
Also, there is the generic Pohlig-Hellman DLP method, which can help if $r$ is of known factorization.Here for we can factor the 89-bit $r/2$ in a breeze. The main DLP reduces to a few easier DLPs in groups of smaller prime order. These can be solved by Baby-step/Giant-step for small ones, or Pollard's rho and it's distributed variant for large ones. A typical factorization of $r$ has a largest prime factor $q$ of multiplicity 1 and much larger than other primes dividing $r$, so that the work is dominated by $\sqrt q$ group operations. Here, it is not told if $p$ was chosen such that $q=r/2$ is prime (which would make Pohlig-Hellman pointless), thus there is a fair chance that $q$ is sizably less than 89-bit, easing attack greatly.
There are even more efficient methods in the case of the multiplicative group $\mathbb Z_p^*$ that we target:
¹ The question's mention of Diffie-Hellman key exchange suggests $p$ is prime. If not, we can factor $p$ into $\prod{p_i}^{\alpha_i}$, solve for $a_i$ the problems $B^{a_i}\equiv C\pmod{{p_i}^{\alpha_i}}$, then use the Chinese Remainder Theorem modulo the ${p_i}^{\alpha_i-1}\,(p_i-1)$ to find a solution $a$.
