Hello, Susie38!
It's a straightforward area problem.
Exactly where does your difficulty lie?
Let $\displaystyle T$ be a triangle with vertices $\displaystyle (0,0),\;(0,c^2),\;(c,c^2)$
and let $\displaystyle R$ be the region between $\displaystyle y = cx$ and $\displaystyle y = x^2,\; c > 0$
Find the ratio $\displaystyle \frac{area\:T}{area\:R}$ Code:

(0,cē)*      * (c,cē)
 /:
 /::*
 /::::
 /::::*
 /:::*
   **     
(0,0)
The base of the triangle is $\displaystyle c$, its height is $\displaystyle c^2$.
. . Its area is: $\displaystyle A_T \:=\:\frac{1}{2}(c)(c^2) \:=\:\frac{1}{2}c^3$
The area of the parabolic segment is:
. . $\displaystyle \displaystyle{A_R \;= \;\int^c_0(cx  x^2)\ dx \;=\;\frac{c}{2}x^2  \frac{1}{3}x^3\bigg^c_0 \;= \;\frac{1}{6}c^3$
Therefore: $\displaystyle \displaystyle{\frac{A_T}{A_R}\;=\;\frac{\frac{1}{2 }c^3}{\frac{1}{6}c^3} \;= \;3$