For those of you who hate math and anyone who practices that heresy, please travel at your own risk, but you might consider that I'm just going to give you a formula for approximating your wing offset, if you want to use offset. It involves multiplying, adding, and dividing. I do an example for you.
For a number of years, Martin Hepperle had a simplified analysis of what he called the lift "eccentricity" across the wing of a control line aircraft. He used elementary calculus, but confined himself to rectangular wings with neither assumed tip losses nor twist. This eccentricity was the distance the center of pressure moves outward from the wing's center, due to increasing air speed outward along the span. He did this by dividing computed rolling moment by total lift. Intuitively I see this as giving a larger eccentricity than would be true for tapered wings with tip losses. Nonetheless, his figures come out smaller than the usual old offsets. I suppose factoring in the line weight would enlarge the figure. Maybe including line weight and taper/tip loss effects would balance each other.
Unfortunately, while Martin's site still has many fascinating things for modelers, I can no longer find this and a related CL derivation there. However, FWIW, below is what I got from that site and my own similar derivation along the same lines. I did not try the derivation for tapered wings, but that shouldn't be really difficult. In fact, it might even be easier to just do it for an ellipse. If you know your line length and span, you can compute the approximate fuselage offset to have equal lift (really lifting moment about the fuselage) on both wings. For my planes, (.25 - .40 power, approx. 50" spans), offsets have usually been around .5" - .75", again not taking into effect the line weight nor the non-linear lift distribution, which would tend to counteract each other.
For instance if I had 60" lines, then l would be about 60' + 1.5' + 2' = 63.5' or about 762". For a 50"-span plane the eccentricity would be...
(50")2/(6 x 762" + 502//(2 x 762")) = 2500 in2/(4572" + 1.64") = 2500 in2/4574 in = .55".
So for this plane, you'd come close by moving the fuselage .55" or about 9/16" outward along the span.
Below is the "formula" as formerly derived on the Hepperle site.