Design a tuned BJT common emitter amplifier with some specifications
I don't think you'll get 34 dB gain with a common emitter amplifier and a load of 600 ohms. I'm not going to go through that process but I will show you what I did regards a common base amplifier because I think that might be your only option: -



The above simulation uses micro-cap 12 and the blue boxes show the DC operating currents and the magenta boxes show the various node voltages. I've used the 2N2222 BJT and the same values for L and C as the question calculated. The input is sourced at V2 and V1 in the spectrum below is the collector.

Here is the spectrum: -



Gain is 33.7 dB at 456 kHz. Gain can be increased to 34 dB by lowering R5 a tad. Transient response looks good: -



The input peak level is 500 mV so it is being over-driven but the tuned circuit does a good job of restoring the sinewave shape.

I don't think you can "adequately" achieve your aims in common emitter but I'm prepared to eat my words and my shirt and hat. 

There will be small-print associated with me eating my "stuff" but I'll hold onto that just in case someone tries to enforce it.

Where the basic problem lies

The big deal is a thing called \$r_E\$ - this resides inside the emitter of the BJT and it limits just how much gain can be achieved in CE or CB. If \$r_E\$ is 26 ohms (for example) then the maximum gain of the circuit is 600 ohm / 26 ohm = 27.3 dB (23.1) and well short of the 34 dB (50) required. So you have to make \$r_E\$ smaller.

To lower \$r_E\$ you need more collector current because: -

$$r_E = \text{26 mV}/ I_C $$ 

So, with Ic = 10 mA, \$r_E\$ is 2.6 ohms. With Ic at 1 mA, \$r_E\$ is 26 ohms and that can never give you a gain of 34 dB (given that the load is 600 ohms).

With your Re (R3) of 4000, you are miles away from getting a quiescent current of circa 10 mA or above. I estimate that you get a quiescent current of 2.16 mA and this make \$r_E\$ = 12 ohm and right at the limit. You need to get heavier handed with the quiescent current or use two transistors.

Good luck and I hope the bias points help.