Answer
a) See the detailed answer below.
b) $6\times10^8 \;\rm V/m$
Work Step by Step
a) We know that the electric field energy density is given by
$$u_E=\dfrac{PE}{\rm Volume}=\frac{1}{2}\epsilon E^2$$
Plugging the known;
$$u_E =\frac{1}{2}\cdot8.85\times10^{-12}\cdot (10^4)^2$$
$$u_E =\color{magenta}{\bf 4.425\times10^{-4}} \;\rm J/m^3\tag 1$$
And we know that the magnetic field energy density is given by
$$u_B=\frac{B^2}{2\mu_0} $$
Plugging the known;
$$u_B=\frac{2^2}{2\cdot 4\pi\times 10^{-7}} $$
$$u_B =\color{magenta}{\bf 1.59\times10^{6}} \;\rm J/m^3\tag 2$$
To compare, divide (2) by (1);
$$\dfrac{u_B }{u_E}=\dfrac{1.59\times10^{6}}{4.425\times10^{-4}}=3.59\times 10^9$$
Therefore,
$$ u_B =\color{magenta}{\bf3.59\times 10^9}\;u_E$$
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b)
Finding the needed magnitude electric field t produce the same energy as the magnetic field of 2T.
$$u_E=u_B$$
Thus,
$$ \frac{1}{2}\epsilon E^2=\frac{B^2}{2\mu_0}$$
Solving for $E$;
$$ E^2=\frac{B^2}{ \mu_0 \epsilon}$$
$$ E =\sqrt{\frac{B^2}{ \mu_0 \epsilon}}= \frac{B}{ \sqrt{\mu_0 \epsilon}}$$
Plugging the known;
$$ E = \frac{2}{ \sqrt{ 4\pi\times10^{-7}\cdot8.85\times10^{-12} }}=5.99\times10^8$$
$$ E =\color{magenta}{\bf6\times10^8}\;\rm V/m$$