This was very straightforwards RSA. n is easily factored (I used the ECM factoring applet [here](https://www.alpertron.com.ar/ECM.HTM). Then just textbook decrypt.
```from Arithmetic import *
p = '338 924256 021210 389725 168429 375903 627261'.replace(' ','')q = '338 924256 021210 389725 168429 375903 627349'.replace(' ','')
p = int(p)q = int(q)
c = 102692755691755898230412269602025019920938225158332080093559205660414585058354
n = p*q
e = 113
phi = (p-1)*(q-1)d = modinv(e,phi)
print hex(pow(c,d,n))[2:-1].decode('hex')```