= Solution
<negative feedback> and the ideal <operational amplifier>'s <virtual short> hold the inverting node at zero potential, with no current entering either input. The parallel feedback <resistor> and <capacitor> have <electrical impedance>
$$
Z_f=\left(R_f^{-1}+i\omega C\right)^{-1}
=\frac{R_f}{1+i\omega R_fC}.
$$
<Kirchhoff's current law> at the inverting node gives $V_{\rm in}/R_i=-V_{\rm out}/Z_f$. For the component values read from the circuit,
$$
G(f)=\frac{V_{\rm out}}{V_{\rm in}}
=-\frac{1000}{1+i\,2\pi f(1\,\mathrm s)},\qquad
\boxed{|G(f)|=\frac{1000}{\sqrt{1+(f/0.15915\,\mathrm{Hz})^2}}}.
$$
This <inverting low-pass amplifier> has a DC gain magnitude of 1000 and <cutoff frequency> $f_c=(2\pi R_fC)^{-1}=0.15915$ Hz. At the corner, $|G|=707.1$. Above it the gain approaches $1/(2\pi fR_iC)=159.15/f$, with $f$ in Hz: a slope of $-1$ on logarithmic axes, or $-20$ dB per decade on a <Bode plot>. At $0.01,1,10,100$ Hz the magnitudes are approximately $998,157.2,15.91,1.592$. These are voltage ratios, not <decibels>.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-1-amplifier-gain.png]
{title=Calculated magnitude response of the inverting low-pass amplifier}
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