This is RSA with a small public exponent. It can be broken by taking the third root of the ciphertext:

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import gmpy
from Crypto.Util.number import long_to_bytes

n = 92164540447138944597127069158431585971338721360079328713704210939368383094265948407248342716209676429509660101179587761913570951794712775006017595393099131542462929920832865544705879355440749903797967940767833598657143883346150948256232023103001435628434505839331854097791025034667912357133996133877280328143
c = 2044619806634581710230401748541393297937319
e = 3

<span>m = gmpy.root(c, 3)[0]
if pow(m, 3, n) == c:
    print long_to_bytes(m)
</span>--------------------------------------------------------
Output: so_low