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
Hint
Starting from lab bhattacharjee's answer, $$V(h)=\dfrac{\pi}3h(45^2-h^2)$$ and you want to maximize $V$.
When is reached an extremum of a function ? So, since you know it, write the corresponding condition and solve for $h$. Plug this value in $V(h)$ and use the second derivative test to confirm that the result corresponds to a maximum.
I am sure that you can take from here.