How to convert from Black body intensity to MJy/sr? How does one convert from  ${\rm ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot Hz^{-1}}$ to ${\rm MJy \cdot sr^{-1}}$
and from ${\rm ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot cm^{-1}}$ to ${\rm MJy \cdot sr^{-1}}$
Part 1
On the one hand, ${\rm 1\, ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot Hz^{-1}} = 10^{-3} {\rm J}/({\rm m^2} \cdot {\rm sr})$ has basic dimensions ${\rm mass} \cdot {\rm time}^{-2} \cdot {\rm angle}^{-2}$ and is a radio brightness.
On the other hand, there is ${\rm 1 MJy}\cdot{\rm sr}^{-1} = 10^{-20} {\rm J}/({\rm m^2} \cdot {\rm sr})$ which has indeed the same basic units. A possible difficulties could be the conversion from Jansky to SI units: $1\, {\rm Jy} = 10^{-26} {\rm W} \cdot {\rm m}^{-2} \cdot {\rm Hz}^{-1}$ and (less likely) the fact that ${\rm M}=10^6$. This said, there is a factor of $10^{-17}$ between the two values.
Part 2
We start with ${\rm 1\, ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot cm^{-1}} = 0.1 {\rm W}/({\rm sr \cdot m^2})/{\rm m} $ which is easily identified by Wolframalpha as unit for a volume scattering emission coefficient. This is not the same as the expression in part 1. However, since you state that it is about Black body intensity, you can follow the argumentation of uhoh in the comments to convert from a wavenumber (given in ${\rm cm}^{-1}$) to a frequency (in ${\rm Hz}$).

However if you'd like to try it yourself, start with $f= c \lambda$, take the derivative to get
$$\frac{{\rm d}f}{{\rm d}\lambda} = -c \frac{1}{\lambda^2} $$
and rearrange to get
$${{\rm d}\lambda} = -c \frac{1}{\lambda^2} {\rm d}\lambda$$
derivative to get
$${\rm d}f = -c \frac{1}{\lambda^2} {\rm d}\lambda$$
and use $c \approx 2.9979\cdot10^{10} {\rm cm}\cdot{\rm s^{−1}}$. Use that to convert your number in ${\rm cm}^{−1}$ to ${\rm Hz}^{−1}$ [...]