= Solution
Assume $A$ is nonempty, so $0<\alpha\leq1$, and put $a=1_A$. By the <Parseval identity on a finite group>, $\sum_r|\widehat a(r)|^2=\alpha$. Select the nonzero frequencies
$$
R=\{r\ne0:|\widehat a(r)|\geq\alpha^{3/2}/2\}.
$$
Then $|R|\alpha^3/4\leq\alpha$, giving \b[$|R|\leq4\alpha^{-2}$]. Outside $R\cup\{0\}$, the fourth Fourier moment has the bound
$$
\sum_{r\notin R\cup\{0\}}|\widehat a(r)|^4
\leq\frac{\alpha^3}{4}\sum_r|\widehat a(r)|^2
=\frac{\alpha^4}{4}.
$$
The fourfold <normalized convolution on a finite group> $H=a*a*1_{-A}*1_{-A}$ has transform $|\widehat a(r)|^4$, so <Fourier inversion on a finite group> gives $H(x)=\sum_r|\widehat a(r)|^4e_N(rx)$. It is real and nonnegative, and $H(x)>0$ implies $x\in2A-2A$ because the convolution counts representations, with a positive normalization.
Use the <Bohr set in phase-distance convention>
$$
B(R,1/10)=\{x:\|rx/N\|_{\mathbb R/\mathbb Z}\leq1/10\text{ for every }r\in R\}.
$$
On this set all retained <additive characters> have positive real part. The zero-frequency contribution is $\widehat a(0)^4=\alpha^4$, and the total absolute contribution of the discarded frequencies is at most $\alpha^4/4$. Hence
$$
H(x)\geq\alpha^4-\alpha^4/4>0\qquad(x\in B(R,1/10)).
$$
Therefore
$$
\boxed{B(R,1/10)\subseteq A+A-A-A,\qquad |R|\leq4/\alpha^2.}
$$
This is an explicit <Bogolyubov lemma>. In the chord-distance convention for a <Bohr set>, the phase-radius $1/10$ corresponds to chord radius $2\sin(\pi/10)>1/10$; alternatively the same positivity proof works directly at chord radius $1/10$. Thus the requested lower bound on width holds in either convention. If $R$ is empty, the <Bohr set> is the whole group. Nonemptiness of $A$ is necessary for the density statement.
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