Answer
Please see the work below.
Work Step by Step
We know that
$v=\sqrt{\frac{\gamma P}{\rho}}$
This simplifies to:
$\gamma=\sqrt{\frac{v^2 \rho}{P}}$
We plug in the known values to obtain:
$\gamma=\frac{(368\frac{m}{s})^2\times 1\frac{Kg}{m^3}}{81000Pa}=1.67=\frac{5}{3}$
