Estimation of pressure and kinetic energy density of stellar interior using kinetic theory of gases
I would like to first point out that the pressure you've determined is too low, most likely because you used only molar mass of protons $\ce{H+}$ instead of average mass for star matter in order to fulfill the condition of plasma quasineutrality:

$$\bar{M} = \frac{M(\ce{H+}) + M(\ce{e-})}{2} = \frac{\pu{(1 + 0) g mol-1}}{2} = \pu{0.5 g mol-1} = \pu{5e-4 kg mol-1} \tag{1.1}$$

Referring to the question from Atkins' Elements of Physical Chemistry [1, p. 40], where this question has been taken from, we obtain necessary data to complete the solution:


  In the Sun [...] the density is 150 times that of liquid water at its centre and comparable to that of water about half-way to its surface


and


  (a) Calculate the pressure half-way to the centre of the Sun, assuming that the interior consists of ionized hydrogen atoms, the temperature is $\pu{3.6 MK}$, and the mass density is $\pu{1.20 g cm−3}$ (slightly higher than the density of water)


the pressure of star plasma (that can be treated as an ideal gas) can be found as following:

$$p = \frac{\rho RT}{\bar{M}} = \frac{\pu{1.2e3 kg m-3} \cdot \pu{8.314 J K-1 mol-1} \cdot \pu{3.6e6 K}}{\pu{5e-4 kg mol-1}} = \pu{7.18e13 Pa} \label{eqn:1.2}\tag{1.2}$$



Now for the second part, kinetic energy density is defined as kinetic energy per volume:

$$\rho_\text{k} = \frac{E_\text{k}}{V} \label{eqn:2.1}\tag{2.1}$$

Kinetic energy for a set of entities (ion, atom, molecule) is

$$E_\text{k} = \frac{Nm\bar{v^2}}{2} \label{eqn:2.2}\tag{2.2}$$

where $N$ is the number of particles, $m$ is the particle's mass, $\bar{v}$ is an average (root mean square) particle velocity. From the kinetic model of an ideal gas, the pressure obtained from \eqref{eqn:1.2} is also

$$p = \frac{Nm\bar{v^2}}{3V} \label{eqn:2.3}\tag{2.3}$$ 

Equation \eqref{eqn:2.3} implies $Nm\bar{v^2} = 3pV$, so kinetic energy from \eqref{eqn:2.2} can be rewritten as $E_\text{k} = \displaystyle\frac{3pV}{2}$. Finally, kinetic energy density, taking \eqref{eqn:2.1} into account, is

$$\rho_\text{k} = \frac{3p}{2} = \frac{3 \cdot \pu{7.18e13 Pa}}{2} = \pu{1.08e14 J m-3} \quad (\pu{1 J} = \pu{1 Pa}\cdot \pu{1 m3})$$

Comparison between kinetic energy densities $\rho_\text{k}(1/2r_\odot)$ (halfway to the Sun's center) and $\rho_\text{k}(\text{atm})$ (Earth's atmosphere) is really a matter of comparing the pressures:

$$\frac{\rho_\text{k}(1/2r_\odot)}{\rho_\text{k}(\text{atm})} = \frac{p(1/2r_\odot)}{p(\text{atm})} = \frac{\pu{7.18e13 Pa}}{\pu{101325 Pa}} = 7.09\cdot10^8$$

References


Atkins, P. W.; De Paula, J. Elements of physical chemistry, 5th ed.; Oxford University Press: Oxford; New York, 2009. ISBN 978-1-4292-1813-9.