Answer
$b_m$ converges to $e^3$.
Work Step by Step
We have
$$
\lim _{m \rightarrow \infty} b_m=\lim _{m \rightarrow \infty}
\left(1+\frac{1}{m}\right)^{3m}
\\
=\left(\lim _{m \rightarrow \infty}
\left(1+\frac{1}{m}\right)^{m}\right)^{3}
=e^3.
$$
Where we used the fact that $\lim _{m \rightarrow \infty}
\left(1+\frac{1}{m}\right)^{m}=e$.
So, $b_m$ converges to $e^3$.
