If force (F), length (L) and time (T) are taken as the fundamental...

d ∝ {F}^{a}{L}^{b}{T}^{c}$</p><p>\left[M{L}^{- 3}\right] = {\left[ML{T}^{- 2}\right]}$^{a}\left[{L}^{b}\right]\left[T{}^{c}\right]

$= \left[{M}^{a}{L}^{a + b}{T}^{- 2a + c}\right]$

comparing :$</p><p>$a = 1$......(i)$

$a + b = - 3$ ....(ii)$</p><p>$- 2a + c = 0$.......(iii)$

$c = 2; b = - 4; ∴ [ d ] = \left[{FL}^{- 4}{T}^{2}\right]</p>$