We are missing e & q as stated by the description and e is under 50. A typical way to solve RSA if we had all the values is with a script like this (Credit: Practical CTF)

e = <value>
p = <value>
q = <value>
n = p*q

phi = (p-1)*(q-1)
d = pow(e, -1, phi)
m = pow(c, d, n)
print(long_to_bytes(m))

Now to obtain q it's pretty simple in this case, we just need to divide n by p which is as simple as...

>>> q = n//p
>>> print(q)
155694605282054119425521475624300384083466178240627480935854427486918260095552504358478493985940907113009654254657147586057507955661767483844303106556370030901659779958849987047274410348258176148315164774766854079870085660650560858927588536846613174753328861645595054240895935716160988886137891728973522588917
>>>

Now we have q, now the last value needed is e. To accomplish this I wrote a python script to bruteforce.

from Crypto.Util.number import long_to_bytes

# our values
c = 7834381455537086069556470828674580173937271064256312815617230923582264273260067511896680320170885743343862894164014864043792487985706669274975353978277862462944814782646749746217479200710773218743409525658958249817055354592575831865920206596021699022326281524908299055420849742148757177093582285192053105592180465511200588277457610702671508363048935552446809692594715753573724603294565113603946429767347234628199252177612170759392617992009035004668823723769453952068616628571785811538463850603629779083508146990944772896050051747702814005551146368404685501836088557927084895438654990821206561992369510510301944786242
p = 95035264145462998106373959950852388512916398417336694051973007035267892127571038290551358518210018988802168144062568058000141570285306734135476955708641860308084865175837570650537276267265396611644179740194499506782555051319215145789689879081854479885459274078337276115880870922739027746148771680782305865397
q = 155694605282054119425521475624300384083466178240627480935854427486918260095552504358478493985940907113009654254657147586057507955661767483844303106556370030901659779958849987047274410348258176148315164774766854079870085660650560858927588536846613174753328861645595054240895935716160988886137891728973522588917

n = p * q
phi = (p - 1) * (q - 1)

# Brute-forcing `e` to find the flag
for e in range(1, 50):
    try:
        d = pow(e, -1, phi) # RSA Magic
        m = pow(c, d, n)
        flag = long_to_bytes(m)

        if b'csd{' in flag:
            print(f"FLAG: {flag.decode()}")
            break
    except ValueError:
        continue

And running it we get...

$ python3 sol.py
FLAG: csd{V3sA_R3sa_RSa?_1D3k}

Flag: csd{V3sA_R3sa_RSa?_1D3k}