First, I’d like to correct one of your assertions: the minor axis length of the ellipse is not equal to the diameter of the horizontal cross section at the point that the plane intersects the cone’s axis. From your diagram you can see that the ellipse’s center is offset from this axis, so you have to measure a chord of the cone that goes through the midpoint of that segment to get the minor axis length.
Quanto has given you an answer to your immediate question about how to compute the major axis length. I’d like to offer an alternative way to find the ellipse’s area that doesn’t require computing its semiaxis lengths directly, nor, for that matter, most of the other measurements in your diagram.
Per the conic section Wikipedia article, an ellipse given by the general conic equation $ax^2+bxy+cy^2+dx+ey+f=0$, or in homogeneous matrix form $\mathbf x^TC\mathbf x=0$, has the canonical-form equation $${x'^2\over-S/(\lambda_1\Delta)}+{y'^2\over-S/(\lambda_2\Delta)}=1,$$ where $S=\det C$, $\lambda_1$ and $\lambda_2$ are the eigenvalues of the upper-left $2\times2$ submatrix of $C$, and $\Delta$ is its determinant, $ac-b^2/4$, which is also equal to $\lambda_1\lambda_2$. The area of the ellipse is therefore equal to $$\pi\sqrt{{S^2\over\Delta^3}} = \pi{\lvert S\rvert\over\Delta^{3/2}}.\tag1$$
In three dimensions, we can similarly represent a quadric surface with an equation in homogeneous coordinates of the form $\mathbf X^TQ\mathbf X=0$, where $Q$ is a symmetric matrix. We can impose a coordinate system on a plane via a $4\times3$ matrix $M$, so that $\mathbf X=M\mathbf x$ is a parameterization of the plane. Substituting into the quadric equation, $(M\mathbf x)^TQ(M\mathbf x)=\mathbf x^T(M^TQM)\mathbf x=0$, which is the equation in this plane coordinate system of the intersection of the plane and quadric—a conic section. So, if you can find suitable matrices $Q$ and $M$ for your cone and plane, you can then use formula (1) to compute the area directly from these matrices.
I’ll assume that by the “height” $z_0$ of the plane you mean the height at which it intersects the cone’s axis (in your diagram, the distance marked as 33.9410). W.l.o.g. we can assume that the cone’s axis is the $z$-axis and that the plane is parallel to the $x$-axis. A convenient choice of $M$ for this plane is $$M = \begin{bmatrix}1&0&0\\0&\sin\theta&0\\0&\cos\theta&z_0\\0&0&1\end{bmatrix},$$ where $0\lt\theta\le\pi/2$ is the angle between the plane and positive $z$-axis. This coordinate system has its origin at the intersection of the plane and cone axis, one axis parallel to the $x$-axis and the other axis perpendicular to this. We ensure that the axis direction vectors in this matrix are orthonormal so that no adjustment to the computed area will be necessary.
Now, for the cone, you’ve written that you have the radius $r$ of the base, but you need one more piece of information to specify the cone, such as its half-aperture angle $\alpha$ or height $h$. It doesn’t matter which one you’ve been given, as the other can be obtained via the identity $\tan\alpha=r/h$. Again w.l.o.g. we’ll assume that the cone’s vertex has a $z$-coordinate of $h\gt0$. An equation of the cone is $$x^2+y^2=(z-h)^2\tan^2\alpha,$$ which corresponds to the matrix $$Q = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&-\tan^2\alpha&h\tan^2\alpha\\0&0&h\tan^2\alpha&-h^2\tan^2\alpha\end{bmatrix},$$ giving $$C = M^TQM = \begin{bmatrix}1&0&0\\0&1-{\cos^2\theta\over\cos^2\alpha}&(h-z_0)\cos\theta\tan^2\alpha\\0&(h-z_0)\cos\theta\tan^2\alpha&-(h-z_0)^2\tan^2\alpha\end{bmatrix}.$$ Applying formula (1) and simplifying produces $${\pi(h-z_0)^2\sin^2\theta\tan^2\alpha \over \left(1-{\cos^2\theta\over\cos^2\alpha}\right)^{3/2}} = {\pi(r-z_0\tan\alpha)^2\sin^2\theta\over\left(1-{\cos^2\theta\over\cos^2\alpha}\right)^{3/2}}.$$ Using $\tan\alpha=r/h$ and $\sec^2\alpha=1+\tan^2\alpha=1+(h/r)^2$, you can convert this expression to use $h$ instead of $\alpha$, if that’s what you’ve been given to start with. If you really do need the axis lengths, you can extract them from $C$ using standard methods.
Plugging in the values from your diagram, this gives an area of approximately $309.443\pi$ for the ellipse. From this, we can compute that the minor axis length is about $29.5137$, somewhat longer than the $29.3726$ that you got by using the diameter of the perpendicular cross-section. These values jibe with those that GeoGebra computed for me. As a sanity check, extracting the axis lengths from $C$ gives $41.9388$ for the major axis length, in good agreement with your diagram.
