Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know, for $RLC$-circuits, that the resonance frequency occurs when
$$X_L=X_C$$
$$2\pi f L=\dfrac{1}{2\pi f C}$$
Hence,
$$ f^2 =\dfrac{1}{4\pi^2 L C}$$
$$ f =\sqrt{\dfrac{1}{4\pi^2 L C}}$$
Plug the known;
$$ f =\sqrt{\dfrac{1}{4\pi^2(1 \times 10^{-3})(1 \times 10^{-6})}}$$
$$f=\color{red}{\bf 5.033\times 10^3}\;\rm Hz$$
$$\omega=2\pi f=\color{red}{\bf 31.622\times 10^3}\;\rm rad/s$$
$$\color{blue}{\bf [b]}$$
First, we need to find the current at the resonance frequency,
$$I=\dfrac{\varepsilon_0}{R}\tag 1$$
So,
$$V_C=IX_C=\dfrac{I}{2\pi f C }$$
Plug from (1),
$$V_C=\dfrac{ \varepsilon_0}{2\pi f C R }$$
Plug the known;
$$V_C= \dfrac{(10)}{2\pi (5.033\times 10^3)(1 \times 10^{-6})(10) }$$
$$V_C=\color{red}{\bf 31.6}\;\rm V$$
And
$$V_R=IR $$
Plug from (1),
$$V_R= \varepsilon_0 =\color{red}{\bf 10}\;\rm V$$
$$\color{blue}{\bf [c]}$$
We know that
$$\varepsilon =V_R+V_L+V_C$$
This means that the net voltage must equal the source voltage and since the capacitor voltage and the resistor voltage are out of phase, the voltage of the inductor at this moment is equal to the negative capacitor voltage;$ V_C=−31.6$ V.
And hence, we still have a net voltage of 10 V.
