It's not clear what "successive remainders" means. There is only one remainder after a division: (dividend)=(quotient)*(divisor)+(remainder). For example, if 175 is the remainder after division then 286 divides into 64239 224 times, remainder 175. The largest remainder is 213, implying that the divisor is bigger than 213. It's clear that the divisor cannot be divided again by the same divisor to yield a remainder in excess of the divisor. In the example 286 will not divide into 224, so 224 would be a second remainder. If the divisor has to be used three times to divide successively into each quotient to obtain each remainder then the divisor cannot exceed 40 (the approximate cube root of 64239) and therefore the remainders cannot exceed 40. See explanations in square brackets below.
The only other interpretation I can find to the question is that in the process of long division, we have the successive remainders as we proceed with the division. But that doesn't seem to work either.
[The numbers a, b and c are integers and b=cx+213; a=bx+114; 64239=ax+175, making ax=64064. Combining these equations we get: a=64064/x=bx+114, so b=64064/x^2-114/x=cx+213, and c=64064/x^3-114/x^2-213/x. Because 64064=ax, we see that 64064 must be divisible by x. 64064 factorises: 2^6*7*11*13. Because the largest remainder is 213, x>213, because x must be bigger than any remainder it creates. But the first term 64064/x^3 can only yield an integer value if x^3<64064, that is, x<40, which contradicts the requirement x>213.]
[If the successive remainders appear in the long division, and the divisor has 3 digits as expected, and three remainders suggest that the other factor also has 3 digits, then if 175 is the first remainder it must be the result of dividing the divisor into 642. Therefore the multiple of the divisor =642-175=467. Since 467 is prime then the first digit of the other factor must be 1. 64239 divides by 467 137 times with 260 as the remainder. This is not one of the three remainders, so 175 cannot be the first remainder. Try 114: 642-114=528=2^4*3*11. We need a factor >213. 264, 352. This time we appear to have partial success, because 176 (less than 213, but half of 352) goes into 64239 364 times with a remainder of 175! So we have two of the three required remainders. Unfortunately the third remainder is 87, so 114 cannot be the first remainder. Finally, try 213: 642-213=429=3*11*13. The only factor bigger than 213 is 429 itself, and the final remainder after dividing 64239 by 429 is 318, not included in the three remainders given. So the assumption that the remainders arise during long division is also false.]
64064=2^6*7*11*13. Leaving aside the powers of 2, and just looking at combinations of 7, 11 and 13, we get 7*11=77; 7*13=91; 11*13=143, and 7*11*13=1001. By progressively doubling these products we can discover factors bigger than 213: 308=4*77; 364=4*91; 286=2*143, etc. We can also take 7, 11 and 13 and multiply them by powers of 2: 224=32*7; 352=32*11; 416=32*13. We can write the factors of 64064 in pairs, starting with a factor a little bigger than 213: (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104), (704,91), (1001,64), (1232,52), (2002,32), (2464,26), (4004,16), (4928,13), (8008,8), (16016,4), (32032,2). These are (a,x) or (x,a) pairs.
If we take 175 as the remainder, and ignore the other two, then we have a choice of factors. The 3-digit factors are (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104). The only pair in which each factor is greater than 213 is (224,286).