I'll help part of the way and you can then finish the solution. The problem is asking you for what divisor choices m is the residual the same on the left and the right sides of the equation.
First, you factor the two numbers into a product of primes (via Fundamental Theorem of Arithmetic):
1848 = (2)3 (3) (7) (11)
1914 = (2) (3) (11) (29)
Since 2, 3 and 11 are common factors to both numbers they will leave a residual of zero if m = 2, 3 or 11.
But the equation may also be true when the residual is 1 or 2 or 3 , etc.
There is no point in trying divisors greater than 29 since that's the largest prime number above. All the prime numbers below that are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. And we've already established that 2, 3 & 11 produce a zero residual.
This is where you come in to find the residuals using the remaining primes and see if any match to add to your solution set.