Question: Design a simple logic inverter with an NPN transitor and a 1k collecter pull-up resistor. Choose the base resistor (E12) as large as possible (but as to ensure that the transistor saturates when the input is at + 5V)

My answer: I collecter = V / R collecter = 5 / 1 = 5 mA

i base = (Vin - 0 .7) / R base

B = i collecter / i base
i collecter = B * i base
i collecter = B * (Vin - 0.7) / R base
R base = i collecter / B * (Vin - 0.7) = 5/ (50 * (5 - 0.7)) = 0.02k

Right answer: With 1k pull-up saturation current will be (5-0) / 1k = 5mA so minimum base current required is 5mA / 50 = 100 micro A. so (5-0.7)R base >= 100 micro A, so R <= 4.3 / 100 micro A = 43k. We much therefore choose the next lower E12 value i.e. 39k.

I dont really understand the answer, could someone please help? My circuit diagram is attached.

 

I don't know who told you that saturation is a forced beta of 50, but it's normally specified as Ic/10.

Since there is a 1k pull-up resistor on the collector, and Vcc=5v, the highest collector current possible (if Vce could be made = to 0v) is (5v-Vce)/1k Ohms, or (5v-0v)/1000 or 5/1000 = 5mA.

The standard calculation for base resistor for a saturated transistor is:
Rbase = (Vin - Vbe) / (Ic/10)
where:
Vin = the voltage applied to the left side of the base resistor
Vbe = the voltage from the base to the emitter; typically anywhere from 0.63v to 0.9v. 0.7v is a common value used.
Ic = desired collector current.
So in your case:
Rbase = (5v - 0.7v)/(5mA/10)
Rbase = 4.3v/0.0005 A
Rbase = 8.6k Ohms. That is not a standard E24 value of resistance. Normally you would look at a chart of standard values, and select the closest resistor.
Here is such a chart:
http://www.logwell.com/tech/components/resistor_values.html
You will see that 8.2k and 9.1k are the closest standard values. In this case, it would be OK to use either.

 

You did mistake in this
R base = i collecter / B * (Vin - 0.7) = 5/ (50 * (5 - 0.7)) = 0.02k

Rb =[ B*(Vin-Vbe)] / Ic = 215/5mA = 43K