= Solution
A mode that crosses during matter domination subsequently grows by
$$
\frac{a(\tau_0)}{a(\tau_H)}
=\left(\frac{\tau_0}{\tau_H}\right)^2.
$$
Its present amplitude is consequently
$$
|\delta(\mathbf k,\tau_0)|
=\left(\frac{\tau_0}{\tau_H}\right)^2
\tau_H^2\widehat A k^{1/2}
=\widehat A\tau_0^2k^{1/2}.
$$
Now
$$
\tau_0^2=D_{\rm eq}\tau_{\rm eq}^2,
\qquad
\tau_{\rm eq}=\frac{2\pi}{k_{\rm eq}c}.
$$
Squaring therefore gives
$$
\boxed{
|\delta(\mathbf k,\tau_0)|^2
=\frac{C}{k_{\rm eq}^4}k,
\qquad
C=\widehat A^2D_{\rm eq}^2\left(\frac{2\pi}{c}\right)^4.}
$$
Since $\tau_H\propto k^{-1}$, modes crossing chronologically between equality and today satisfy $k_0\leq k\leq k_{\rm eq}$. The reverse ordering printed with the target formula is inconsistent with the stated definitions of $k_0$ and $k_{\rm eq}$; it does not affect the spectrum calculation.
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