Explanation

Let $h \propto G^{x}L^{y}E^{z}$ $\left\lbrack ML^{2}T^{- 1} \right\rbrack \propto \left\lbrack M^{- 1}L^{3}T^{- 2} \right\rbrack^{x}\left\lbrack ML^{2}T^{- 1} \right\rbrack^{y}\left\lbrack ML^{2}T^{- 2} \right\rbrack^{z}$ $\left\lbrack ML^{2}T^{- 1} \right\rbrack = k\left\lbrack M^{- 1}L^{3}T^{- 2} \right\rbrack^{x}\left\lbrack ML^{2}T^{- 1} \right\rbrack^{y}\left\lbrack ML^{2}T^{- 2} \right\rbrack^{z}$ Comparing the powers, we get $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 = - x + y + z\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 = 3x + 2y + 2z\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 1 = - 2x - y - 2z\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$ On solving Eqs. (i), (ii) and (iii), we get $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0$ $\therefore$ Gravitational constant has no dimensions