= Solution
Multiply by the $(m,n)$ sine product and integrate over the square. <Fourier orthogonality> gives squared norm $\ell^2/4$, and thus
$$
c_{mn}=\frac4{\ell^2}\left(\int_0^\ell\sin\frac{m\pi x}{\ell}dx\right)\left(\int_0^\ell\sin\frac{n\pi y}{\ell}dy\right)
=\frac{4[1-(-1)^m][1-(-1)^n]}{\pi^2mn}.
$$
Therefore
$$
\boxed{c_{mn}=\begin{cases}16/(\pi^2mn),&m,n\text{ both odd},\\0,&\text{otherwise}.\end{cases}}
$$
This <double Fourier sine series> equals one at interior points. At boundary points every summand vanishes, consistent with the endpoint value of the sine extension.
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