Question: [![enter image description here](https://i.stack.imgur.com/qe9JA.png)](https://i.stack.imgur.com/qe9JA.png)

I know when it is centered at 0, Vpeak/sqrt(2) gives the RMS.

When I do FFT on LTspice, it gives a peak at -17dB which is equal to 0.141254

[![enter image description here](https://i.stack.imgur.com/0mP38.png)](https://i.stack.imgur.com/0mP38.png)

On the FFT page, program says that it shows RMS values. But, I don't think the RMS of this wave is 0.14 W. You are also welcome to answer for this question involving LTspice.

However, main question is : I actually don't think this question is specific to power calculation, anyway, **what is the method to calculate RMS of such sine waves?**

Answer: 
As the signal is not totally recognizable in the first figure, I'll assume your sine wave equation to be \$ f(t) = 0.25 + 0.25\,sin(5000\,\pi\,t)\$ and \$f(t)\$ has Watt unit.

Thus, the RMS value is given by \$\sqrt{\dfrac{1}{T}\int\_{t\_0}^{t\_0+T} f(t)^2dt}\$ being T the period of the function, which I have assumed to be \$\approx .4\,ms\$. Thus, \$T = .4\,ms\$ and as the function has its peak at \$\approx .5\,W\$ I have thus reached the equation given above. Figuring out the integral value we end up with :

\$\sqrt{\dfrac{1}{0.0004}\int\_{0}^{0.0004}(0.25 + 0.25\,sin(5000\,\pi\,t))^2dt} = 306.18\,m\,W\$ as the approximate RMS value.
The RMS value you are seeing is a combination of the DC and AC components. Your signal is the sum of those two components and the RMS value is taken **of the sum**.

If you want only the RMS value of the AC portion, i.e., the sinusoidal portion, then using the SPICE controls to subtract the DC component and then take the RMS value of that result, which will now be only the AC portion of the signal.

Make sense?