by substituting the following dimensions :

$[G] = [{M^{ - 1}}{L^3}{T^{ - 2}}],\,[c] = [L{T^{ - 1}}],[g] = [L{T^{ - 2}}]$

$[p] = [M{L^{ - 1}}{T^{ - 2}}]$

and by comparing the powers of both sides we can get $x = 0,\,y = 2,\,z = - 1$

$\therefore $ $[G] \propto {c^0}{g^2}{p^{ - 1}}$