Answer
a) $ 3.19\times10^{-2 }\;\rm H$
b) $3.15\times10^{-2}\;\rm A$
c) $1.584\times10^{-5 }\;\rm J$
Work Step by Step
a) We know that the inductance is given by
$$f_0=\dfrac{1}{2\pi \sqrt{LC}}$$
Thus,
$$(2\pi f_0)^2=\dfrac{1}{ LC }$$
$$L=\dfrac{1}{4\pi^2f_0^2 C}$$
Plugging the known;
$$L=\dfrac{1}{4\pi^2\cdot (19\times10^3)^2 \cdot 2200\times10^{-12}}$$
$$L=\color{red}{\bf 3.19\times10^{-2}}\;\rm H$$
------------------------------------------------------------------
b) We can assume that the initial energy inside the capacitor is completely transformed into the maximum energy in the magnetic field at some moment.
Thus,
$$\frac{1}{2}CV_0^2=\frac{1}{2}LI^2_{\rm max}\tag 1$$
$$ CV_0^2= LI^2_{\rm max}$$
Solving for $I_{\rm max}$;
$$ \dfrac{CV_0^2}{ L}=I^2_{\rm max}$$
$$ I_{\rm max}=\sqrt{\dfrac{CV_0^2}{ L}}$$
Plugging the known;
$$ I_{\rm max}=\sqrt{\dfrac{2200\times10^{-12}\cdot 120^2}{ 3.19\times10^{-2}}}=$$
$$ I_{\rm max}=\color{red}{\bf 3.15\times10^{-2}}\;\rm A$$
------------------------------------------------------------------
c) The maximum energy stored in the magnetic field of the inductor is given by
$$U_{L,max}=\frac{1}{2}LI^2_{\rm max}$$
which is, as from (1), equal to
$$U_{L,max}=\frac{1}{2}CV_0^2=\frac{1}{2}\cdot 2200\times10^{-12}\cdot 120^2 $$
$$ U_{L,max} =\color{red}{\bf1.584\times10^{-5}}\;\rm J$$
