If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional formula for mass will be

The correct option is V⁻²F⁰E.

To find the dimensional formula for mass ($M$) in terms of velocity ($V$), force ($F$), and energy ($E$) as fundamental units, we can express mass as a product of powers of these quantities:
$M=k{V}^a{F}^b{E}^c$
where $k$ is a dimensionless constant, and $a$, $b$, and $c$ are the exponents we need to determine.

First, let's write down the dimensional formulas of the fundamental quantities in the standard MLT system:
1. Mass: $\left[M\right]=M^1{L}^0{T}^0$
2. Velocity: $\left[V\right]=L{T}^{-1}$
3. Force: $\left[F\right]=ML{T}^{-2}$
4. Energy: $\left[E\right]=M{L}^2{T}^{-2}$

Substitute these dimensional formulas into the relation:
$\left[M\right]={\left[V\right]}^a{\left[F\right]}^b{\left[E\right]}^c$
$M^1{L}^0{T}^0={\left(L{T}^{-1}\right)}^a{\left(ML{T}^{-2}\right)}^b{\left(M{L}^2{T}^{-2}\right)}^c$
Simplifying the right-hand side by grouping the powers of M, L, and T:
$M^1{L}^0{T}^0=M^{b+c}{L}^{a+b+2c}{T}^{-a-2b-2c}$

By comparing the exponents of M, L, and T on both sides, we get three simultaneous equations:
For M: $b+c=1$ --- (Equation 1)
For L: $a+b+2c=0$ --- (Equation 2)
For T: $-a-2b-2c=0$ --- (Equation 3)

Let's solve these equations:
From Equation 3, we can express $a$ in terms of $b$ and $c$:
$a=-2b-2c$
Substitute this expression for $a$ into Equation 2:
$(-2b-2c)+b+2c=0$
$-b=0\Rightarrow b=0$

Substitute $b=0$ into Equation 1:
$0+c=1\Rightarrow c=1$

Now, substitute the values of $b$ and $c$ back to find $a$:
$a=-2\left(0\right)-2\left(1\right)=-2$

Therefore, the values are:
$a=-2$
$b=0$
$c=1$

Substituting these values back into the dimensional relation gives:
$\left[\text{Mass}\right]=V^{-2}{F}^{0}E$