The dimensional formula for inductance is:

Hint: Using the formulae of energy stored in inductor - $E=\frac{1}{2} L i$.

Step (1) - Calculation of the dimension formula of inductor since, we know that -

Magnetic energy store in inductor $=E=\frac{1}{2} L i^{2}$ 801

$\begin{array}{l} \operatorname{Dim}\left(\frac{1}{2} L(2)=\operatorname{Dim}(E)\right. \\ \operatorname{Dim}\left(\frac{1}{2} L i 2\right)=\left[M L^{2} T^{-2}\right] \\ \operatorname{Dim}(l) \times \operatorname{Dim}\left(l^{2}\right)=\left[M L^{2} T^{-2}\right] \\ \left.\operatorname{Dim}(L) \times[A)^{2}\right]=\left[M L^{2} T^{-2}\right] \\ \operatorname{Dim}(l)=\left[M L^{2} T^{-2} A^{-2}\right] \end{array}$

$\rightarrow$ Hence, option-A is correct