D. 50! is divisible by all the prime numbers less than 50, so you should recognize that the smallest prime number that's not a factor of 50! is 53 (remember, 51 may "look prime" but it's divisible by 3). But the question specifically says that x cannot be prime - so how can you take 53 and make it 'not prime' but 'still small'? Multiply it by the smallest possible factor that will accomplish that: 2. That gives you 106, a number that's clearly not prime (it's even) but that has as its only factors 1, 2, 53, and 106. That's as small a non-prime number as you can get that's not a factor of 50!, so that's x. And the sum of its factors is 162.

While your initial temptation might be to assume that the smallest such integer must be greater than 50!, you should catch yourself: that first integer above 50! is 50! + 1, and its factors must include 50! + 1 and 1. While that matches choice E, remember this - if those are indeed the only factors, then x would be prime (only factors: itself and 1) and that would violate the terms of the question. If that was your first hunch, catching that mistake is an entry point to do a little work with the significantly-smaller numbers in choices A through D.