The first equation can be rearranged like so:
$2N\log(0.2\pi/W) = \log((1/0.89125)^2-1)$
and a similar thing can be done for the second equation. Perform a division to cancel out the $N$, leading to
$\frac{\log(0.2\pi/W)}{\log(0.3\pi/W)} = \frac{\log((1/0.89125)^2-1)}{\log((1/0.17783)^2-1)}$
or
$\frac{\log(0.2\pi)-\log\,W}{\log(0.3\pi)-\log\,W} = \frac{\log((1/0.89125)^2-1)}{\log((1/0.17783)^2-1)}$
Solving for $\log\,W$, and then $W$, should be a snap. Once you have the value of $W$, substitute into any of the two original equations, or the equation like the first one I gave in this answer to solve for $N$.