If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be :

Applying dimensional analysis
$\mathrm{S}\propto {\mathrm{E}}^{\mathrm{a}}{\mathrm{V}}^{\mathrm{b}}{\mathrm{T}}^{\mathrm{c}}$
$$\begin{gathered} {\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^{-2}= \mathrm{k} {\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]}^{\mathrm{a}} {\left[{\mathrm{L}}^1 {\mathrm{T}}^{-1}\right]}^{\mathrm{b}}{\left[{\mathrm{T}}^1\right]}^{\mathrm{c}} \\ {\mathrm{M}}^1{\mathrm{L}}^0 {\mathrm{T}}^{-2} = \mathrm{k}\left[{\mathrm{M}}^{\mathrm{a}}{\mathrm{L}}^{2\mathrm{a}}{\mathrm{T}}^{-2\mathrm{a}}\right]{\mathrm{L}}^{\mathrm{b}}{\mathrm{T}}^{-\mathrm{b}}{\mathrm{T}}^{-\mathrm{c}} \end{gathered}$$
Comparison
$$\begin{gathered} \mathrm{a}=1 \left|\begin{array}{c}2\mathrm{a}+\mathrm{b}= 0 \\ \mathrm{b}=-2\end{array}\right| \\ -2=-2\mathrm{a}-\mathrm{b}+\mathrm{c} \\ -2=-2\left(1\right)+2+\mathrm{c} \\ \mathrm{c}=-2 \end{gathered}$$
So, the dimensional formula for surface tension will be[E¹V^-2T^-2]
Alternate solution : 
$$\begin{gathered} \mathrm{Surface} \mathrm{Tension} =\frac{\mathrm{Surface} \mathrm{energy}}{\mathrm{Area}} \\ [\mathrm{Surface} \mathrm{tension}] =\frac{\left[\mathrm{E}\right]}{{\left[\mathrm{V}.\mathrm{T}\right]}^2}= \left[{\mathrm{EV}}^{-2}{\mathrm{T}}^{-2}\right] \end{gathered}$$