Given: Gravitational constant (G), Planck's constant $(h)$ and speed of light $(c)$ are fundamental units.

To find: The dimensions of time

Dimensional formula of G:

$\lbrack G\rbrack = \left\lbrack M^{- 1}{\text{\ }L}^{3}{\text{\ }T}^{- 2} \right\rbrack,\ \left( F = G\frac{m_{1}m_{2}}{r^{2}} \right)$

Dimensional formula of $h:\$

$\lbrack h\rbrack = \left\lbrack M^{1}{\text{\ }L}^{2}{\text{\ }T}^{- 1} \right\rbrack,\ \left( \lambda = \frac{h}{p} \right)\text{~}$

$\text{Dimensional\ formula\ of~}c\text{:\ }$

$\lbrack c\rbrack = \left\lbrack L^{1}{\text{\ }T}^{- 1} \right\rbrack$

Let

$\lbrack T\rbrack = (G)^{x}(h)^{y}(c)^{z}$

Putting values of $G,h$ and $L$ from equations (i), (ii) and (iii) in equation (iv):

$\lbrack T\rbrack = \left\lbrack M^{- 1}L^{3}T^{- 2} \right\rbrack^{x}\left\lbrack M^{1}L^{2}T^{- 1} \right\rbrack^{y}\left\lbrack L^{1}T^{- 1} \right\rbrack^{z}$

$\lbrack M^{0}L^{0}T^{1}\rbrack = \lbrack M\rbrack^{- x + y}\lbrack L\rbrack^{3x + 2y + z}\lbrack T\rbrack^{- 2x - y - z}$

Comparing the exponents:

$\begin{matrix} - x + y = 0 \\ 3x + 2y + z = 0 \\ - 2x - y - z = 1 \\ \end{matrix}$

Solving

$x = \frac{1}{2},$

$y = \frac{1}{2}\text{~and~}$

$z = \frac{- 5}{2}$

$\begin{matrix} t & \ = G^{\frac{1}{2}}h^{\frac{1}{2}}c^{\frac{- 5}{2}} \\ \text{~or~}\ t & \ \propto \sqrt{\frac{Gh}{c^{5}}} \\ \end{matrix}$