= Solution
A <flat torus> is the quotient $\mathbb R^d/\Lambda$, where $\Lambda$ is a full-rank <Euclidean lattice>, with the <Riemannian metric> induced from the <Euclidean metric>. Translation by a lattice vector is a <Riemannian isometry>, so this metric is well defined and has zero curvature. Its volume is the covolume of $\Lambda$.
The <dual lattice> is $\Lambda^\vee=\{w:\langle w,v\rangle\in\mathbb Z\text{ for all }v\in\Lambda\}$. For $w\in\Lambda^\vee$, the function $e_w(x)=\exp(2\pi i\langle w,x\rangle)$ descends to the <flat torus>, and direct differentiation gives
$$
\Delta e_w=4\pi^2|w|^2e_w.
$$
Conversely, write $\Lambda=B\mathbb Z^d$ and $x=By$. Ordinary <Fourier series> on $\mathbb R^d/\mathbb Z^d$ have frequencies $n\in\mathbb Z^d$, corresponding to $w=B^{-\mathsf T}n\in\Lambda^\vee$. After normalization they form a complete <orthonormal basis> in $L^2$, and the <Fourier coefficients> of a smooth function decay rapidly. Applying the <Laplace-Beltrami operator> term by term shows that a smooth <eigenfunction> with <eigenvalue> $\lambda$ has nonzero coefficients only where $4\pi^2|w|^2=\lambda$. Thus the <spectrum of a flat torus> is
$$
\boxed{\operatorname{Spec}(\mathbb R^d/\Lambda)=\{4\pi^2|w|^2:w\in\Lambda^\vee\},\qquad
m(\lambda)=\#\{w\in\Lambda^\vee:4\pi^2|w|^2=\lambda\}.}
$$
The formula counts multiplicity over either the real or complex numbers: the two frequencies $w,-w$ give the real sine and cosine functions.
For the rigidity assertion in dimension two, it suffices to reconstruct a rank-two <Euclidean lattice> $L=\Lambda^\vee$ from its vector-length multiset. The multiset first determines its covolume $A$. Indeed, a bounded fundamental parallelogram, and comparison of the cells meeting a disk with slightly larger and smaller disks, give the <lattice-point asymptotics by fundamental cells>
$$
N_L(R)=\#\{v\in L:|v|\leq R\}=\frac{\pi R^2}{A}+O(R).
$$
The <spectrum> determines $N_L(R)$, including multiplicities, so it determines $A$.
Let $a$ be the smallest nonzero vector length, and choose $v\in L$ with $|v|=a$. This vector is primitive: $v=ku$ with $|k|>1$ would contradict minimality. From the length multiset subtract the two vectors $kv,-kv$ at every positive length $ka$, for $k=1,2,\ldots$. The shortest remaining length $b$ is exactly the length of a shortest vector $w\notin\mathbb Zv$. This subtraction is legitimate even if several directions have length $a$: it subtracts just two copies, and then $b=a$. It also proves that $b$ is independent of our choice of shortest direction. Replace $w$ by $w-kv$ to arrange
$$
|v\cdot w|\leq a^2/2.
$$
The replacement stays outside $\mathbb Zv$ and cannot be shorter than $b$, so a reduced choice still has length $b$.
The <shortest-independent-vector basis lemma> says that $v,w$ are a basis of $L$. To prove it, choose coordinates with $v=(a,0)$. Because $v$ is primitive, $L/\mathbb Zv\cong\mathbb Z$; let $h>0$ be the smallest positive vertical coordinate in $L$. Write $w=(x,mh)$, with $m\geq1$ after changing its sign. If $m\geq2$, a vector $u$ of vertical coordinate $h$, reduced horizontally modulo $v$, has
$$
|u|^2\leq a^2/4+h^2\leq(a^2+m^2h^2)/4\leq(a^2+b^2)/4\leq b^2/2<b^2.
$$
This contradicts the definition of $b$, since $u\notin\mathbb Zv$. Hence $m=1$ and $v,w$ generate $L$.
The <Gram matrix> of this basis is $\left(\begin{smallmatrix}a^2&c\\c&b^2\end{smallmatrix}\right)$, with $c=v\cdot w$, and its determinant is $A^2$. Therefore
$$
c^2=a^2b^2-A^2.
$$
Changing the sign of one basis vector allows $c\geq0$. The numbers $a,b,A$, all audible from the <spectrum>, consequently determine the <Gram matrix>, and hence $L$, up to an <orthogonal transformation>. Taking <dual lattices> gives the same conclusion for $\Lambda$. \b[Isospectral flat two-dimensional tori are isometric.] The argument uses the special basis property of rank-two <Euclidean lattices>, so it does not claim higher-dimensional rigidity.
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