Answer
(a) The dimensions of $V$ are $[L^3]$
(b) The correct interpretation for $V$ would be volume.
Work Step by Step
(a) A force has the unit newtons which is $kg~m~s^{-2}$. In dimensional analysis, this would be expressed as $[M~L~T^{-2}]$.
Let $u$ be the dimensions of $V$ in the equation $F_B = \rho~g~V$. We can find the dimensions of $V$.
$[M~L^{-3}]~[L~T^{-2}]~u = [M~L~T^{-2}]$
$[M~L^{-2}~T^{-2}]~u = [M~L~T^{-2}]$
$[L^{-2}]~u = [L]$
$u = \frac{[L]}{[L^{-2}]}$
$u = [L^3]$
The dimensions of $V$ are $[L^3]$.
(b) Since volume has the dimensions $[L^3]$, the correct interpretation for $V$ would be volume.
