An efficient way to do it is to switch to polar coordinates. We have $x^2+y^2=r^2$, so our integral is equal to $\int_{\theta=0}^{2\pi}\left(\int_{r=0}^1 (cr^2)r\,dr\right)d\theta.$. The integration is exceptionally easy. Then, as you did, you will be setting the result equal to $1$ to evaluate $c$.
Your procedure will also succeed, with somewhat more work. A small change needs to be made, for correctness.
If one does it more or less your way, the inner integral needs to be from $x=-\sqrt{1-y^2}$ to $x=\sqrt{1-y^2}$. Out of habit, other things being equal, I slightly prefer integrating first with respect to $y$, from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$.
I would suggest exploiting symmetry . Integrate over the first quadrant, and multiply the result by $4$.