As we know, surface tension,
$S=\frac{\text { force }}{\text { length }}=\frac{ F }{ l }$
Hence, its SI unit is $\frac{ N }{ m }$.
So, option (a) is unit of surface tension.
Again, option (b) $\frac{\text { joule }}{(\text { metre })^{2}} \cong \frac{\text { energy }}{(\text { length })^{2}}$
$=\frac{\text { force } \times \text { length }}{(\text { length })^{2}}=\frac{\text { force }}{\text { length }}=\frac{ F }{ l }$
In terms of SI unit, it will be $\frac{ N }{ m }$, hence it is also
equal to surface tension.
Again, option (c) $\frac{ kg }{ s ^{2}} \approx \frac{\text { mass }}{(\text { time })^{2}}$
But $\frac{\text { energy }}{(\text { length })^{2}} =\frac{ F \times s }{ s ^{2}}=\frac{ m \times a \times s }{ s ^{2}}$
$= m \times \frac{\text { distance }}{(\text { time })^{2}} \times \frac{1}{\text { distance }}=\frac{ kg }{ s ^{2}}=\frac{ F }{ l }$
Hence, option (c) also equivalent to surface tension.
But, option (d) $=\frac{\text { watt }}{\text { metre }}=\frac{\text { power }}{\text { mass }}$
$=\frac{\text { force } \times \text { distance }}{\text { mass }}=\frac{ N - m }{ kg }$
cannot be equal to surface tension.