What would be an appropriate value for \$V_{CE}\$ in the absence of a signal?
The ideal value of \$V_{CE}\$ is one that gives the maximum amount of room for the output signal to swing. So for instance, if \$V_{CE}= 18V\$, then the voltage can only go up by 2V before it reaches the supply rail and begins to clip. A similar thing happens if \$V_{CE}=2V\$, except that the voltage now can only go down by \$2V-V_{CEsat} \approx 1.5V\$. The ideal voltage for \$V_{CE}\$ is halfway between the positive and negative supplies, as that gives the maximum amount of room for the amplified signal to be.
Calculate the base current \$I_B\$ and the base resistor \$R_B\$
Your calculation for \$I_B\$ looks good, but your calculation for \$R_B\$ could add the a diode drop to the base voltage, though in a real design this wouldn't matter that much.
Calculate the value of \$R_C\$ to give your chosen value of \$V_{CE}\$
This looks fine, as long as you change \$V_{CE}\$ as described above
What is the maximum input signal
The problem doesn't give any description of the input source, which is rather important if you want to use a constant diode drop model for \$V_{BE}\$. Assuming an answer in amps is what is wanted, calculate the base current needed to make\$V_{CE} = V_{CEsat} \approx 0.3V\$.
Is this a good amplifier?
No. Realistically, you need emitter degeneration on a practical amplifier, as it makes the gain of the transistor depend more on \$\frac{R_C}{R_E}\$ and less on \$\beta\$. Additionally, you want a voltage divider type biasing network instead of the single resistor, again so it is easier to bias with a variable \$\beta\$. The stage will work fine in the ideal case presented here, but with the variations in beta among transistors in the real world, the circuit becomes much less practical.
