Consequences of Flipping a bit in an RSA public key
First note that $e=17=10001_2$ and thus that $e'=11001_2=25$ with $e'$ being the one with the flipped bit. Now note that $M^{e'}\equiv C_1\pmod n$ and $M^{e}\equiv C_2\pmod n$.

Next note how $C_2/C_1\equiv M^{25}/M^{17}\equiv M^8 \pmod n$.
Now compute $M^{17}/M^8\equiv M^{9}\pmod n$ and continue this.
$M^{9}/M^{8}=M \pmod n$ and you are done.

Note that all divisions here are in $\mathbb Z^*_n$, that is, they are multiplications with the multiplicative inverses $\bmod n$.