Expression for time in terms of $G$ (universal gravitational constant), $h$ (Planck constant) and $c$ (speed of light) is proportional to:
(A) $\sqrt {\dfrac{{h{c^5}}}{G}} $
(B) $\sqrt {\dfrac{{{c^3}}}{{Gh}}} $
(C) $\sqrt {\dfrac{{Gh}}{{{c^5}}}} $
(D) $\sqrt {\dfrac{{Gh}}{{{c^3}}}} $

Hint We need to express the time in terms of $G$, $h$ and $c$, using the process to convert a physical quantity from one system to the other. We should use the expressions for the gravitation, planck’s constant, and the speed of light to convert their physical system.

Formula used In the solution we will be using the following formula
$\Rightarrow F = \dfrac{{GMm}}{{{R^2}}}$
$\Rightarrow h = \dfrac{E}{v}$
$\Rightarrow c = 3 \times {10^8}m/s$
Where, $G$ is the universal gravitational constant, $h$ is the planck’s constant, $c$ is the speed of light, $M$ and $m$ are masses of earth and the body, $E$ is energy, $R$ is the distance between the earth and the body, and $v$ is the frequency.

Complete step by step answer
Let us consider the universal gravitational constant as $G$, the Planck constant as $h$, and the speed of light as $c$.
We can write the expressions for the above constants as,
For gravitational constant,
$\Rightarrow F = \dfrac{{GMm}}{{{R^2}}}$
From here we can write the value of the gravitational constant as,
$\Rightarrow G = \dfrac{{F{R^2}}}{{Mm}}$
For planck’s constant can be written as,
$\Rightarrow h = \dfrac{E}{v}$
For the speed of light we have,
$\Rightarrow c = 3 \times {10^8}m/s$
To convert a physical quantity from one system to other, we have to use this expression,
$\Rightarrow {n_1}\left[ {{u_1}} \right] = {n_2}\left[ {{u_2}} \right]$
We can write this as,
$\Rightarrow {n_2} = {n_1}{\left[ {\dfrac{{{M_1}}}{{{M_2}}}} \right]^a}{\left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right]^b}{\left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]^c}$
We can write the dimensional formula for the constants as,
$\Rightarrow G = [{M^{ - 1}}{L^3}{T^{ - 2}}]$
$\Rightarrow h = [M{L^2}{T^{ - 1}}]$
$\Rightarrow C = [L{T^{ - 1}}]$
We can write that,
$\Rightarrow t \propto {G^x}{h^y}{C^z}$
Then, when we substitute the quantities in the above expression, we get,
$\Rightarrow [T] = {[{M^{ - 1}}{L^3}{T^{ - 2}}]^x}{[M{L^2}{T^{ - 1}}]^y}{[L{T^{ - 1}}]^z}$
On comparing the powers of M, L and T we can write,
$\Rightarrow - x + y = 0 $
$\Rightarrow x = y $
And,
$\Rightarrow 3x + 2y + z = 0 $
$\Rightarrow 5x + z = 0 $
And,
$\Rightarrow - 2x - y - z = 1 $
$\Rightarrow 3x + z = - 1 $
After solving the above two equations,
We get,
$\Rightarrow x = y = \dfrac{1}{2}$ and
$\Rightarrow z = \dfrac{{ - 5}}{2}$
Thus, when we substitute the values of x, y, z, we get,
$\Rightarrow t \propto \sqrt {\dfrac{{Gh}}{{{c^5}}}} $
Hence, the correct answer is option (C).

Note
We should know the SI units of the physical quantities to convert it from one system to the other, so that we can change the system to find out the relation. While converting, make sure to cancel out the proportionality by using variables.