From the dimensional consideration, which of the following equation is correct

The correct option is T = 2π√(R³/GM).

To determine the correct formula using dimensional analysis, let us analyze the dimensions of each variable involved in the equation:
1. Time period, $T$ has the dimension of time:
$\left[T\right]=M^0{L}^0{T}^1$

2. Radius, $R$ has the dimension of length:
$\left[R\right]=L$

3. Mass, $M$ has the dimension of mass:
$\left[M\right]=M$

4. Universal Gravitational Constant, $G$ can be derived from Newton's Law of Gravitation, $F=\frac{G{m}_1{m}_2}{r^2}$:
$\left[G\right]=\frac{\left[F\right]\left[r^2\right]}{\left[m_1\right]\left[m_2\right]}=\frac{\left(ML{T}^{-2}\right)\left(L^2\right)}{M^2}=M^{-1}{L}^3{T}^{-2}$

Now, let us find the dimensions of the term $GM$:
$\left[GM\right]=\left(M^{-1}{L}^3{T}^{-2}\right)\left(M\right)=L^3{T}^{-2}$

Next, let us check the dimensions of the expression on the right-hand side of the correct option, $\sqrt{\frac{R^3}{GM}}$ (noting that the constant $2\pi$ is dimensionless):
$\left[\sqrt{\frac{R^3}{GM}}\right]={\left(\frac{L^3}{L^3{T}^{-2}}\right)}^{\frac{1}{2}}={\left(T^2\right)}^{\frac{1}{2}}=T$

Since the dimensions of the left-hand side ($T$) match the dimensions of the right-hand side ($T$), the equation $T=2\pi \sqrt{\frac{R^3}{GM}}$ is dimensionally correct.