Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
For an $RC$ circuit, the capacitor voltage is given by
$$V_C=\dfrac{\varepsilon_0X_C}{\sqrt{R^2+X_C^2}}$$
where $X_C=1/\omega C$;
$$V_C=\dfrac{\varepsilon_0}{\omega C}\dfrac{1}{\sqrt{R^2+\dfrac{1}{\omega^2 C^2}}}$$
$$V_C= \dfrac{\varepsilon_0}{\sqrt{\omega^2 C^2R^2+1}}$$
where $\omega=2\pi f$;
$$V_C= \dfrac{\varepsilon_0}{\sqrt{ 4\pi^2 f^2 C^2R^2+1}}$$
Plug the known;
$$\boxed{V_C= \dfrac{(10)}{\sqrt{ 4\pi^2 f^2 (10^{-6})^2(16)^2+1}}}$$
$\bullet$ At $f=1$ kHz,
$$V_C= \dfrac{(10)}{\sqrt{ 4\pi^2 (1\times 10^3)^2 (10^{-6})^2(16)^2+1}}=\color{red}{\bf 9.95}\;\rm V$$
$\bullet$ At $f=3$ kHz,
$$V_C= \dfrac{(10)}{\sqrt{ 4\pi^2 (3\times 10^3)^2 (10^{-6})^2(16)^2+1}}=\color{red}{\bf 9.57}\;\rm V$$
$\bullet$ At $f=10$ kHz,
$$V_C= \dfrac{(10)}{\sqrt{ 4\pi^2 (10\times 10^3)^2 (10^{-6})^2(16)^2+1}}=\color{red}{\bf 7.05}\;\rm V$$
$\bullet$ At $f=30$ kHz,
$$V_C= \dfrac{(10)}{\sqrt{ 4\pi^2 (30\times 10^3)^2 (10^{-6})^2(16)^2+1}}=\color{red}{\bf 3.15}\;\rm V$$
$\bullet$ At $f=100$ kHz,
$$V_C= \dfrac{(10)}{\sqrt{ 4\pi^2 (100\times 10^3)^2 (10^{-6})^2(16)^2+1}}=\color{red}{\bf 0.99}\;\rm V$$
$$\color{blue}{\bf [b]}$$
Plugging the five points we got, to draw the $V_C$ versus $f$ graph.
$\bullet$ $\rm (1\;kHz,9.95\;V)$
$\bullet$ $\rm (3\;kHz,9.57\;V)$
$\bullet$ $\rm (10\;kHz,7.05\;V)$
$\bullet$ $\rm (30\;kHz,3.15\;V)$
$\bullet$ $\rm (100\;kHz,0.99\;V)$
