= Solution
A suitable <electrical circuit> has $R$ in series with the input and a shunt branch consisting of $C$ in parallel with the lossy <inductor> $R_L+i\omega L$; take the output <voltage> across that shunt branch. This is the middle arrangement in the circuit figure. In terms of its <complex impedance>,
$$
Z_p=\left(i\omega C+\frac1{R_L+i\omega L}\right)^{-1},\qquad H(\omega)=\frac{U_2}{U_1}=\frac{Z_p}{R+Z_p}.
$$
At <direct current> the <capacitor> is an <open circuit> and the <inductor> contributes only $R_L$, so $H(0)=R_L/(R+R_L)=0.01$. Therefore
$$
\boxed{R=99\,\Omega.}
$$
When $R_L$ is neglected, $Z_p=i\omega L/(1-\omega^2LC)$. At the <antiresonance> $\omega_0^2LC=1$ its <electrical admittance> vanishes, so $|H|$ reaches its maximum value of one. Consequently
$$
\boxed{C=\frac1{L\omega_0^2}=1.78\times10^{-3}\,\mathrm F.}
$$
This value uses precisely the requested lossless approximation; keeping $R_L$ changes both the peak position and its height. A <transfer function> peak and one DC measurement do not uniquely identify every possible <two-port network>, but the displayed arrangement meets both observations to the stipulated approximation. Also, the angle brackets in the source must denote AC <wave amplitudes> or <root mean square> values: the literal time averages of sinusoidal <voltages> are zero, whose ratio would be undefined.
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