Putting the dimensions in the above relation

$[M{L^2}{T^{ - 1}}] = k{[L{T^{ - 1}}]^x}{[L{T^{ - 2}}]^y}{[ML{T^{ - 2}}]^z}$

$⇒$ $[M{L^2}{T^{ - 1}}] = k[{M^z}{L^{x + y + z}}{T^{ - x - 2y - 2z}}]$

Comparing the powers of $M,\,L$ and $T$

$z = 1$ …$(i)$

$x + y + z = 2$ …$(ii)$

$ - x - 2y - 2z = - 1$ …$(iii)$

On solving $(i)$, $(ii)$ and $(iii)$ $x = 3,\,y = - 2,\,z = 1$

So dimension of $L$ in terms of $v,\,A$ and $f$
$[L] = [F{v^3}{A^{ - 2}}]$