Calculate ozone layer thickness at a given temperature and pressure
Final conditions ($\pu{1.0 atm}$) give a hint at what altitude the thickness should be estimated – at the Earth's surface. Let's denote the unknown thickness as $h_x$, radius of the Earth as $r_\mathrm{E} = \pu{6371 km}$; also take $h_1 = \pu{15 km}$ and $h_2 = \pu{35 km}$ and put these dimensions on a rough 2D sketch:



Fig. 1. Schematic representation of two ozone layers at given altitudes.

To find $h_x$, we can use the volume of the new hypothetical ozone layer $V_x$ as the difference in volumes of two spheres with corresponding radii:

$$V_x = \frac{4 \pi (r_\mathrm{E} + h_x)^3}{3} - \frac{4 \pi r_\mathrm{E}^3}{3} \tag{1}$$

and express $h_x$ from (1) as follows:

$$h_x = \left(\frac{3V_x}{4 \pi} +r_\mathrm{E}^3\right)^{\frac{1}{3}} - r_\mathrm{E} \tag{2}$$

From ideal gas law for a relatively thin gas layer without temperature and pressure gradients:

$$V_x = \frac{nRT_x}{p_x} \tag{3}$$

where $n$ is the amount of gas in ozone layer, $T_x = \pu{20 ^\circ C}$ and $p_x = \pu{1 atm}$ (new conditions). Let's also denote ozone concentration as $N = \pu{3.0e13 cm-3}$, then total amount of ozone is to be found via initial volume $V_0$ and Avogadro constant $N_\mathrm{A}$ as

$$n = V_0 \times \frac{N}{N_\mathrm{A}} \tag{4}$$

Initial volume can also be found from Fig. 1 using the difference in volumes of the corresponding spheres with radii $r_\mathrm{E} + h_1$ and $r_\mathrm{E} + h_2$:

$$V_0 =  \frac{4 \pi (r_\mathrm{E} + h_2)^3}{3} - \frac{4 \pi (r_\mathrm{E} + h_1)^3}{3} = \frac{4 \pi}{3}((r_\mathrm{E} + h_2)^3 - (r_\mathrm{E} + h_1)^3) \tag{5}$$

Now all we have to do is carefully plug in the values with the corresponding units carefully and find all the unknown variables. From (5):

$$V_0 = \frac{4 \pi}{3}((\pu{6371 km} + \pu{35 km})^3 - (\pu{6371 km} + \pu{15 km})^3) = \pu{1.03e10 km3} = \pu{1.03e25 cm3} \tag{6}$$

Now let's find the amount of ozone using $V_0$ and equation (4):

$$n = \pu{3.0e13 cm-3} \times \frac{\pu{1.03e25 cm3}}{\pu{6.023e23 mol-1}} = \pu{5.12e14 mol} \tag{7}$$

and, subsequently, the unknown volume $V_x$ using (3):

$$V_x = \frac{\pu{5.12e14 mol} \cdot \pu{8.314 J mol-1 K-1} \cdot \pu{293 K}}{\pu{101325 Pa}} = \pu{1.23e13 m3} = \pu{1.23e4 km3} \tag{8}$$

Finally, plugging all the values in (2), we can find the thickness:

$$h_x = \left(\frac{3 \cdot \pu{1.23e4 km3}}{4 \pi} + (\pu{6371 km})^3\right)^{\frac{1}{3}} - \pu{6371 km} = \pu{2.4e-5 km} = \pu{24 mm} \tag{9}$$

The result is somewhat close to the fact posted by NASA:


  Around the world, the ozone layer averages about 3 millimeters (1/8 inch) thick, approximately the same as two pennies stacked one on top of the other.