In a new system of units, $1$ unit of force is $100 N$, $1$ unit of velocity is $2 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ and $1$ unit of time is $4$ sec. Then, density of water $1000 {\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}^3=\_\_\_\_\_\_\_\_\_\_$ units in new system.

$$\begin{gathered} 1F_0=100 N \\ 1v=2\raisebox{1ex}{$ m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ 1 T=4 sec \end{gathered}$$

$$\begin{gathered} M{L}^{-3}={\left[ML{T}^{-2}\right]}^a{\left[L{T}^{-1}\right]}^b{\left[T\right]}^c \\ 1=a \end{gathered}$$

$$\begin{gathered} -3=a+b \\ \Rightarrow b=-4 \end{gathered}$$

$$\begin{gathered} -2a-b+c=0 \\ \Rightarrow c=2a+b=-2 \end{gathered}$$

$$\begin{gathered} \Rightarrow 1000 \raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.=1000 N{\left(\frac{m}{s}\right)}^{-4}{\left(sec\right)}^{-2} \\ =1000\times \frac{1}{100}\times {\left(\frac{1}{2}\right)}^{-4}\times {\left(\frac{1}{4}\right)}^{-2} units \\ =10\times 2^4\times 4^2 \\ =2560 units \end{gathered}$$