check the picture below.
[tex]\bf A=2\pi pw\quad
\begin{cases}
w=7\\
A=693\pi
\end{cases}\implies 693\pi =2\pi p7\implies \cfrac{693\pi }{14\pi }=p
\\\\\\
49.5=p\\\\
-------------------------------\\\\
r=p-\cfrac{w}{2}\implies r=49.5-3.5\implies r=46[/tex]
now, that's just the radius of the pool, the diameter would be twice as much, or 2r.
