Calculating minimum resistance such that Ic is less than or equal to the maximum Ic in BJT
On my calculator, 5/0.3 is 16.7, not 18; but other than that you are right. 

But you are right that this is a very low value. The resistor will be generating \$(0.3\ \rm{A})^2(18\ \Omega)\$, or 1.5 W of heat, for no particular purpose, unless the resistor represents a heater element.

Also, this circuit will not actually put the BJT into saturation unless the input signal goes above the 5 V power rail, which is a somewhat uncommon circumstance. On the other hand, if you drive the input too far above 5 V, you will forward bias the b-c junction, resulting in high current draw from the input. This makes the circuit more difficult to control correctly, even if you do have a way to drive the input above the supply voltage.

You should probably give your answer to this homework, then forget you ever saw this circuit. If you want to switch power to a resistive load, don't do it this way. Google "low-side switch" for the most common way to do it.