The slant heights intersect the ellipse base at diametrically opposite points. I'm going to assume that these points of intersection are at the top and bottom of the ellipse so that the line between them is the major axis. The minor axis is the width, and you noted that the width is a quarter the length of the height. So, we call the width $2a$ and the height $8a$. These are the lengths of the minor axis and the major axis of our ellipse, respectively.
Now, the perimeter of an ellipse is $C = 4(4a)E(e)$, where $e$ is the eccentricity of the ellipse and $E$ is the complete elliptic integral of the second kind. The eccentricity of our ellipse is,
$$e = \sqrt{1 - \frac{a^2}{4a^2}} = \frac{\sqrt{3}}{2}$$
Now, we can approximate $E(e) \approx 1.1315$ (I used Wolfram Alpha). Thus, you can calculate $a$ from the equation $C = 4(4a)E(e)$ since you know the perimeter.
Once you've calculated $E$, then you have have to determine the height of the cone from the tip to base. Using the major axis and the two slant heights, you have a triangle. You know all three side lengths, and you should be able to set up a system of equations to solve for the height.
Finally, once you have the height and the value of $a$ in our ellipse, you can find the area. As Narasimham said, you can still use the formula $V = \frac{1}{3}Bh$ where $B$ is the area of the base and $h$ the height. The area of your ellipse will be $B = \pi(a)(4a) = 4\pi a^2$. You know the height, so you can calculate the volume.
Best Answer
If the condition were merely that the path cannot go downhill, to find the shortest path you could unroll the cone to a circular wedge with radius $60$ and vertex angle $\frac23\pi.$ The shortest path would run along a straight segment from one end of the radius-$60$ arc to a tangent point on a concentric arc of radius $50$ and then follow the radius-$50$ arc to the other end. If $L$ is the length of this path, then $$L=\sqrt{60^2-50^2} + 50\left(\frac23\pi - \arccos\frac56\right).$$
Here's a rough sketch of the unrolled cone and the key features of the shortest never-downhill path:
Angle $\angle AOB = \frac23\pi,$ angle $\angle AOT = \arccos\frac56,$ the line $AT$ is tangent to the arc $BT$ at $T,$ and the path consists of the segment $AT$ and the arc $BT.$
Intuitively, since we can never get closer than distance $50$ from $O$ (since otherwise we'd have to go downhill to get to $B$), the path is constrained to the region between the two arcs. If we imagine running a string from $A$ to $B$ and then pull the string as tight as possible, all while staying within that region, the string will follow the path shown.
Any other path that goes around the cone from $A$ to $B$ will either go downhill at some point or will be longer than $L.$ But you can construct a path that is uphill everywhere and whose length is as close to $L$ as you want simply by replacing the arc of radius $50$ by a suitable spiral. For example, construct a spiral path from $B$ to a point $C$ on $OA$ such that $OC = 50.000001,$ then construct a line through $A$ tangent to that spiral at a point $T'$; the path is the segment $AT'$ plus the part of the spiral from $T'$ to $B,$ all of it strictly uphill. To make an even shorter path, set $OC = 50.000000001$ instead of $50.000001.$ The only limit on how short you can make this path is that if you reduce $OC$ to $50$ exactly, then you have a path of exactly length $L,$ but then the path is not strictly uphill the whole way.
Hence the number you are seeking, the length of the shortest uphill path around the cone from $A$ to $B,$ is the smallest number greater than $L.$