Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the peak inductor voltage is given by
$$V_L=IX_L$$
where $X_L=\omega L$, and $I=\varepsilon_0/Z$
$$V_L=\dfrac{\varepsilon_0\omega L}{Z}\tag 1$$
where, in $RLC$ circuits, $Z=\sqrt{(X_L-X_C)^2+R^2}$ where $X_C=q/\omega C$
$$V_L=\dfrac{\varepsilon_0\omega L}{\sqrt{(\omega L-\frac{1}{\omega C})^2+R^2}}$$
$$V_L=\dfrac{\varepsilon_0\omega L}{\left[ \frac{L^2}{\omega^2}(\omega^2 -\frac{1}{LC})^2+R^2\right]^{1/2}}$$
where $1/LC=\omega_0^2$;
$$V_L=\dfrac{\varepsilon_0\omega L}{ \left[ \frac{L^2}{\omega^2}(\omega^2 -\omega_0^2)^2+R^2\right]^{1/2}}$$
$$V_L=\dfrac{\varepsilon_0\omega^2 L}{ \left[ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2\right]^{1/2}}$$
Maximizing the inductor voltage means that $dV_L/d\omega=0$.
So
$$ \dfrac{d}{d\omega} \dfrac{\varepsilon_0\omega^2 L}{ \left[ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2\right]^{1/2}}=0$$
$$ \dfrac{2\varepsilon_0\omega L\left[ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2\right]^{1/2}-\frac{1}{2} \left[ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2\right]^{-1/2}[4\omega L^2(\omega^2-\omega_0^2)+2\omega R^2](\varepsilon_0\omega^2 L)}{ \left[ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2\right] }=0$$
Hence,
$$2\varepsilon_0\omega L\left[ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2\right]^{1/2}=\frac{1}{2} \left[ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2\right]^{-1/2}[4\omega L^2(\omega^2-\omega_0^2)+2\omega R^2](\varepsilon_0\omega^2 L) $$
$$2 \left[ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2\right]^{1/2}= \left[ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2\right]^{-1/2}[2 L^2(\omega^2-\omega_0^2)+ R^2] \omega^2 $$
$$2 L^2 (\omega^2 -\omega_0^2)^2+2\omega^2R^2 = 2 L^2\omega^2(\omega^2-\omega_0^2)+\omega^2 R^2 $$
$$2 L^2 (\omega^4 -2\omega^2 \omega_0^2+\omega_0^4) + \omega^2R^2 = 2 L^2\omega^4 -2 L^2\omega^2\omega_0^2 $$
$$2 L^2 \omega^4 -4L^2\omega^2 \omega_0^2+2L^2\omega_0^4 + \omega^2R^2 = 2 L^2\omega^4 -2 L^2\omega^2\omega_0^2 $$
$$ 2L^2\omega_0^4 + \omega^2R^2 = 2 L^2\omega^2\omega_0^2 $$
$$ \omega^2R^2 - 2 L^2\omega^2\omega_0^2 =-2L^2\omega_0^4 $$
$$ \omega^2(R^2 - 2 L^2 \omega_0^2) =-2L^2\omega_0^4 $$
$$ \omega^2 =\dfrac{-2L^2\omega_0^4 }{R^2 - 2 L^2 \omega_0^2}$$
since $\omega_0^2=1/LC$, $\omega_0^4=1/L^2C^2$
$$ \omega^2 =\dfrac{-2L^2 }{L^2C^2[R^2 - 2 L^2 \omega_0^2]}$$
$$ \omega^2 =\dfrac{-2 }{ C^2[R^2 - 2 L^2 \omega_0^2]}$$
$$ \omega^2 =\dfrac{1 }{ -\frac{1}{2}C^2 R^2 + C^2 L^2 \omega_0^2}$$
$$ \omega^2 =\dfrac{1 }{ -\frac{1}{2}C^2 R^2 + C^2 L^2 \frac{1}{LC}}$$
$$ \omega =\sqrt{\dfrac{1 }{ -\frac{1}{2}C^2 R^2 + C L }}$$
$$ \omega = \left[ C L -\frac{1}{2}C^2 R^2 \right]^{-1/2}$$
$$ \boxed{\omega_L = \left[ \dfrac{1}{\omega_0^2} -\frac{1}{2}C^2 R^2 \right]^{-1/2}}$$
$$\color{blue}{\bf [b]}$$
$\Rightarrow$ when $\omega=\omega_0$, then $Z=R$
Plug into (1),
$$V_L=\dfrac{\varepsilon_0\omega L}{Z}= \dfrac{\varepsilon_0\omega_0 L}{R}$$
So,
$$V_L= \dfrac{\varepsilon_0 L}{R\sqrt{LC}}$$
$$V_L= \dfrac{\varepsilon_0 }{R}\sqrt{\dfrac{L}{C}}$$
Plug the given;
$$V_L= \dfrac{(10)}{(1)}\sqrt{\dfrac{(1\times 10^{-6})}{(1\times 10^{-6})}}$$
$$V_L=\color{red}{\bf 10}\;\rm V$$
$\Rightarrow$ when $\omega=\omega_L$, then $Z=\sqrt{(\omega_L L-\dfrac{1}{\omega_L C})^2+R^2}$
Plug into (1),
$$V_L=\dfrac{\varepsilon_0\omega_L L}{\sqrt{(\omega_L L-\dfrac{1}{\omega_L C})^2+R^2}}\tag 2 $$
Plug the given into the boxed formula above to find $\omega_L$,
$$ \omega_L = \left[ \dfrac{1}{\omega_0^2} -\frac{1}{2}C^2 R^2 \right]^{-1/2} = \left[ LC -\frac{1}{2}C^2 R^2 \right]^{-1/2} $$
$$ \omega_L = \left[ (10^{-6})^2 -\frac{1}{2}(10^{-6})^2 (1)^2 \right]^{-1/2} $$
$$\omega_L =\bf 1.414\times 10^6\;\rm rad/s$$
Plug the known into (2),
$$V_L=\dfrac{(10)(1.414\times 10^6)( 10^{-6})}{\sqrt{\left((1.414\times 10^6)( 10^{-6})-\dfrac{1}{(1.414\times 10^6)( 10^{-6})}\right)^2+1^2}} $$
$$V_L=\color{red}{\bf 11.55}\;\rm V$$
