Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$For an $RC$ circuit, the resistor voltage is given by
$$V_R=\dfrac{\varepsilon_0R}{\sqrt{R^2+X_C^2}}$$
where $X_C=1/\omega C$;
$$V_R=\dfrac{\varepsilon_0R}{\sqrt{R^2+\dfrac{1}{\omega^2 C^2}}}$$
$$V_R=\dfrac{\varepsilon_0 }{\sqrt{1+\dfrac{1}{\omega^2 R^2C^2}}}$$
where $\omega=2\pi f$;
$$V_R=\dfrac{\varepsilon_0 }{\sqrt{1+\dfrac{1}{4\pi^2 f^2 R^2C^2}}}$$
Plug the known;
$$\boxed{V_R=\dfrac{(10)}{\sqrt{1+\dfrac{1}{4\pi^2 f^2 (100)^2(1.6\times 10^{-6})^2}}}}$$
$\bullet$ At $f=100$ Hz,
$$V_R=\dfrac{(10)}{\sqrt{1+\dfrac{1}{4\pi^2 (100)^2 (100)^2(1.6\times 10^{-6})^2}}}=\color{red}{\bf 1.00}\;\rm V$$
$\bullet$ At $f=300$ Hz,
$$V_R=\dfrac{(10)}{\sqrt{1+\dfrac{1}{4\pi^2 (300)^2 (100)^2(1.6\times 10^{-6})^2}}}=\color{red}{\bf 2.89}\;\rm V$$
$\bullet$ At $f=1000$ Hz,
$$V_R=\dfrac{(10)}{\sqrt{1+\dfrac{1}{4\pi^2 (1000)^2 (100)^2(1.6\times 10^{-6})^2}}}=\color{red}{\bf 7.09}\;\rm V$$
$\bullet$ At $f=3000$ Hz,
$$V_R=\dfrac{(10)}{\sqrt{1+\dfrac{1}{4\pi^2 (3000)^2 (100)^2(1.6\times 10^{-6})^2}}}=\color{red}{\bf 9.49}\;\rm V$$
$\bullet$ At $f=10,000$ Hz,
$$V_R=\dfrac{(10)}{\sqrt{1+\dfrac{1}{4\pi^2 (10,000)^2 (100)^2(1.6\times 10^{-6})^2}}}=\color{red}{\bf 9.95}\;\rm V$$
$$\color{blue}{\bf [b]}$$
Plugging the five points we got, to draw the $V_R$ versus $f$ graph.
$\bullet$ $\rm (100\;Hz,1.00\;V)$
$\bullet$ $\rm (300\;Hz,2.89\;V)$
$\bullet$ $\rm (1000\;Hz,7.09\;V)$
$\bullet$ $\rm (3000\;Hz,9.49\;V)$
$\bullet$ $\rm (10000\;Hz,9.95\;V)$
