Process equation is $p{V}^2=\beta$  $...\left(\mathrm{i}\right)$

As gas is ideal, it obeys gas equation, $pV=nRT$   $...\left(\mathrm{iii}\right)$

From Eqs. $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, gives

$\left(nRT\right)·V=\beta$

Here, $n=0.02$ moles,

$R=8.31 {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}, T=20°\mathrm{C}+273=293 \mathrm{K}$

and $V=1500 {\mathrm{cm}}^3=1.5\times {10}^{-3} {\mathrm{m}}^3$

$\therefore \beta =0.02\times 8.31\times 293\times 1.5\times {10}^{-3}$

$=7.3\times {10}^{-2} \mathrm{Pa}-{\mathrm{m}}^6$

$\approx 7.5\times {10}^{-2} \mathrm{Pa}-{\mathrm{m}}^6$