Let $a=3.05,\ b=2.23.$ Then a parametric equation for the ellipse is $x=a\cos t,\ y=b \sin t.$ When $t=0$ the point is at $(a,0)=(3.05,0)$, the starting point of the arc on the ellipse whose length you seek. Now it's important to realize that the parameter $t$ is not the central angle, so you need to get the value of $t$ which corresponds to the top end of your arc. At that end you have $y/x=\tan 50$ (degrees). And in terms of $t$ you have $y/x=(b/a)\tan t$. Solving for $t$ then gives $$t=t_1=\arctan \left( \frac{a}{b}\tan 50 \right).$$

[note I'd suggest using radians here, replacing the $50$ by $5\pi/18.$]

For the arclength use the general formula of integrating $\sqrt{x'^2+y'^2}$ for $t$ in the desired range. In your case $x'=-a \sin t,\ y'=b \cos t$, so that you are integrating $$\sqrt{a^2 \sin^2t+b^2 \cos^2t}$$ with respect to $t$ from $0$ to the above $t_1$. There not being a simple closed form for the antiderivative (it's an "elliptic integral), the simplest approach now would be to do the integral numerically. This seems the more appropriate in your problem as you only know $a,b$ to two decimals, apparently.

* When I did this numerically on maple I got about $2.531419$ for the arclength.

You can compute this as

$$d=b\,E\bigl(\tan^{-1}(a/b\,\tan(\theta))\,\big|\,1-(a/b)^2\bigr)$$

using the incomplete elliptic integral of the second kind ^[1] $E(\varphi\,|\,m)$. In Mathematica-Syntax (and suitable for Wolfram Alpha ^[2]) this can be written as

2.23*EllipticE[ArcTan[3.05/2.23*Tan[50°]],1-(3.05/2.23)^2]

I adapted this from this post ^[3] which investigates the converse problem (given arc length, find angle) but along the way treats this direction of the problem as well. As noted there, this angle conversion will only work for the first and last quadrant. Otherwise, either adjust the angle or look at that post for an alternative formula to use in its place.

With a few more digits of precision, the answer is returned as $2.5314195265536624417$ which essentially matches both the other answers here. Of course, printing that many digits in the answer is very bad style if the input is only given to two decimals. It does show that the numerical integration by Jyrki is a bit less precise than what coffeemath did, but even he should theoretically have rounded in the other direction.

Note that the formula above only works for $-\frac\pi2<\theta<\frac\pi2$. For $\frac\pi2<\theta<\pi$ the result of $\tan^{-1}$ will represent $\theta-\pi$ so to correct for that you can add $\pi$ to that result. Similar for $-\pi<\theta<-\frac\pi2$ where you need to subtract $\pi$ from the $\tan^{-1}$ result. Generally speaking you want the first input to the $E$ function to be an angle in the same quadrant as $\theta$, adding integer multiples of $\pi$ as necessary.

Giving a Mathematica calculation. Same result as coffeemath (+1)

In[1]:= ArcTan[3.05*Tan[5Pi/18]/2.23]
Out[1]= 1.02051
In[2]:= x=3.05 Cos[t];
In[3]:= y=2.23 Sin[t];
In[4]:= NIntegrate[Sqrt[D[x,t]^2+D[y,t]^2],{t,0,1.02051}]
Out[4]= 2.53143

I believe I found this asymptotic approximation from Jacobi's elliptical integral of the second kind some time ago. It's not precise, but converges exactly for degenerate cases of $b=0$ (lines) and $b=a$ (circles). The infinite series methods are ideal where precision is desired. I offer this only as a curiosity. The last term, recently added, contributes little, but ensures convergence for circles. Assume $0\le b\le a$. Then $$ s\approx 2 \pi \sqrt{a b}+(4 a)^{\frac{a-b}{a+b}} -a^{b-a}. $$