Answer
(a) $U = 10.93 \times 10^{3} \,\text{J}$
(b) Yes, it does.
(c) $u = 28.6 \times 10^{6} \mathrm{~J/m^3}$
(d) $u = 52 \times 10^{3}\mathrm{~J/m^3}$
Work Step by Step
(a) The energy storage is given by
\begin{align*}
U &= \dfrac{1}{2} C V^2\\
& = \dfrac{1}{2} (3 \times 10^{3} \,\text{F}) (2.7 2. \,\text{V})^2 \\
&= \boxed{10.93 \times 10^{3} \,\text{J}}
\end{align*}
(b) This energy is $10.93 \times 10^{3} \,\text{J} = 10.93 \times 10^{3} \,\text{W.s}$. So let us convert is from $\mathrm{W\cdot s}$ to $\mathrm{Wh}$ by dividing our reslut by 3600 s/h
$$U = \dfrac{10.93 \times 10^{3} \mathrm{~W\cdot s}}{3600 \,\text{s/h}} \simeq 3 \,\text{Wh}$$
Hence, the answer is yes, it does.
(c) The energy density $u$ is the energy store $U$ divided by the volume of the cylinder capacitor
\begin{align*}
u &= \dfrac{U}{V}\\
&= \dfrac{U}{(\pi r^2) L}\\
&= \dfrac{10.93 \times 10^{3} \,\text{J}}{(\pi (0.03 \,\text{m})^2) (0.135 \,\text{m})}\\
&= \boxed{28.6 \times 10^{6} \mathrm{~J/m^3}}
\end{align*}
(d) Energy is the amount of work stored in the capacitor and the energy density $u$ is related to the electric field by
$$u = \dfrac{1}{2} \epsilon_o K E^2$$
$\epsilon_o$ is the permittivity of the free space and equals $8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2}$. Substitute to get $u$
\begin{align*}
u &= \dfrac{1}{2} \epsilon_o K E^2\\
&= \dfrac{1}{2} (8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2})(3.3) (6 \times 10^{7} \,\text{V/m})^2 \\
& =\boxed{ 52 \times 10^{3}\mathrm{~J/m^3}}
\end{align*}
