= Solution
The original diagram places the output across the $10\,\Omega$ <resistor>, in series with an ideal $1\,\mathrm{mH}$ <inductor>. Treat the voltages as <phasors> with time dependence $e^{i\omega t}$. The <electrical impedances> are $Z_R=R$ and $Z_L=i\omega L$, so the same <electric current> flows through both:
$$
I=\frac{V_0}{R+i\omega L},\qquad
\boxed{H(\omega)=\frac{V_1}{V_0}=\frac{R}{R+i\omega L},\qquad
|H|=\frac{1}{\sqrt{1+(\omega L/R)^2}}.}
$$
This is an <RL low-pass filter>, with corner <angular frequency>
$$
\omega_c=\frac RL=\frac{10\,\Omega}{10^{-3}\,\mathrm H}
=10^4\,\mathrm{rad\,s^{-1}}.
$$
The <frequency response> tends to one for $\omega\ll\omega_c$, has magnitude $1/\sqrt2$ at the corner, and tends to $\omega_c/\omega$ for $\omega\gg\omega_c$. The requested sketch uses a logarithmic horizontal axis and a linear voltage-gain axis:
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019-ia-physics-q2-rl-response.png]
{title=Voltage gain of the original RL circuit, with signal, corner and noise frequencies marked}
At the two specified <angular frequencies>,
$$
|H(10^3)|=\frac1{\sqrt{1.01}}=0.995037,\qquad
|H(10^5)|=\frac1{\sqrt{101}}=0.0995037.
$$
The <relative attenuation by a linear filter> is the ratio of these gains:
$$
\boxed{\frac{(V_{\rm noise}/V_{\rm signal})_{\rm out}}
{(V_{\rm noise}/V_{\rm signal})_{\rm in}}
=\frac{|H(10^5)|}{|H(10^3)|}=0.100.}
$$
Thus the noise <amplitude> is \b[attenuated by a factor of 10 relative to the signal]. A squared-voltage ratio would decrease by a factor of $100$; the question's voltage transfer function gives the amplitude comparison.
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