How much Energy is Stored in this Capacitor?
The LC tank, that you've drawn, will oscillate / ring, and as you've stated in your question, "This is a real world circuit with losses", it will come to a still. During the oscillations the same current will go back and forth several times through the same losses, aka resistances. 

So having an inductor will drain more power than having no inductor. The total power lost (aka energy) will be greater with the inductor than without the inductor. I'm implying the same thing with different words. 

If the LC tank is critical damped or overdamped then the current will only pass through the resistances once. If it's underdamped however.. Then you get oscillation / ringing.

Hmm, this is my gut feeling though, so I may be wrong. I don't feel that I have time to put up the math to prove myself wrong or right. It's not worth the effort. 

When it has stopped oscillating it will reach a steady state and no more power will be lost. And then it will reach your strange 12.33 V. 

And to reach the climax of your "question". The capacitance hasn't changed, the voltage hasn't changed, the equation for the energy stored in a capacitor hasn't changed. It doesn't matter how it was put in there. After the steady state, when everything has come to a still, the energy stored in the capacitor can be calculated, which you've done beforehand. So I'll take your word on it being "76.014uJ".



Your question is weird and I feel that it doesn't matter what I write, will the correct answer give you anything meaningful? 

When I feel that I'm becoming misunderstood, I take a step back and ask myself "Who is my audience?"
"Do they know what I know?"
"Do they need to know something else to grasp the information?"

In other words, The problem is 99% of the time with me. Perhaps.. you need to reword your question so you don't have to make 3 nearly identical questions within 24h. 



Try #2
LC tank

At T=0 when the switch closes, the following things will happen: 

The capacitor, together with the inductor, is like a loaded spring. But it will "decompress" and surge current and start oscillate. I'm explaining it weirdly, but it's my view of it. And according to the link, the energy stored in an LC tank which is what you got in that setup, is equal to \$\frac{1}{2}CV^2 + \frac{1}{2}LI^2\$. At T=0 then the current will be 0. If we know the voltage and the capacitance, which we do, then we can calculate it. In other words, the total energy in the LC tank will be what you had in your question. "76.014uJ". However, this energy will be moving around in the circuit. At some point in time the majority of it will be in the capacitor, some other point in time in the inductor. 

Since the oscillations are going in both directions, then... at some point in time the capacitor will be fully charged in one direction, such as having 12.33V across it's terminals. Fully loaded. Then in some other point in time it's voltage will be entirely inverted and it will have the same amount of energy. Such as having -12.33V across its terminals. This is probably how you are seeing a value above 50%. The equation is \$\frac{1}{2}CV^2\$, if V is -12.33 or 12.33 then the energy stored will be the same, right? But if you calculate some weird energy difference then you might see 99.8% because it's swinging to the top of some cycle and then to the bottom of some other cycle. It's hard to explain but I can see that it's possible that you've come to terms that it's 99.8%. While in reality it's 49.9%. 



After trying it out in a simulator I've come to terms that at T=0, both the AC and DC energy are put into the LC tank. After the AC part has decayed, then the DC part is left which is equivalent to \$\frac{1}{2}CV^2\$ J. In other words, at T=0, \$CV^2\$ is inside of the LC tank in the form of AC + DC, the AC will have an amplitude that is half of the DC, so the \$V_{PP}\$ of the AC is the same value as the DC level. 

Solving it for DC first, you get what you had in your question, "76.014uJ", solving it for AC after a tiny amount of time has elapsed so everything is stored inside of the capacitor and none is inside of the inductor, you get "76.014uJ" as well. Then you put the AC component + the DC component together and you get an answer that is close to that of laptop2d, his has decayed a little bit which is why it's slightly less than 100%. I used superposition in order to solve this problem. I'm not a clever man. It took me 3 tries. 

So the 50% part I spoke about earlier that was "swooshing" around, that's not a lie, but it was also a 50% DC component laying on top of everything which confused me and most likely you. 

TLDR; At T=0 the system contains 2×76.014uJ≃152µJ. After the AC component is gone, the DC component, 76.014uJ is left inside of the capacitor.