= Solution
With the symmetric Fourier normalization displayed in the question, the <creation and annihilation operators> obey the <canonical commutation relation>
$$
\boxed{[\hat a_{\mathbf k},\hat a_{\mathbf k'}^\dagger]
=\delta^{(3)}(\mathbf k-\mathbf k')},
\qquad
[\hat a_{\mathbf k},\hat a_{\mathbf k'}]
=[\hat a_{\mathbf k}^\dagger,\hat a_{\mathbf k'}^\dagger]=0.
$$
Since $\delta\hat\phi=\hat f/a$, its vacuum <two-point correlation function> is
$$
\langle0|\delta\hat\phi(\tau,\mathbf x)
\delta\hat\phi(\tau,\mathbf x+\mathbf r)|0\rangle
=\int\frac{d^3k}{(2\pi)^3}
\frac{|f_k(\tau)|^2}{a^2}e^{-i\mathbf k\cdot\mathbf r}.
$$
Comparison with the definition in the question gives
$$
\Delta_{\delta\phi}^2(k,\tau)
=\frac{k^3}{2\pi^2}\frac{|f_k|^2}{a^2}.
$$
Using $|f_k|^2=(1+1/(k^2\tau^2))/(2k)$ and $a=-1/(H\tau)$,
$$
\boxed{\Delta_{\delta\phi}^2(k,\tau)
=\left(\frac H{2\pi}\right)^2(1+k^2\tau^2)}.
$$
On <superhorizon scales>, $k\ll aH$ or $|k\tau|\ll1$, so the <scale-invariant inflationary power spectrum> freezes to
$$
\boxed{\Delta_{\delta\phi}^2=\left(\frac H{2\pi}\right)^2}.
$$
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