$\frac{(4.53-3.07)^2}{(3.07-2.15)(2.15-4.53)} + \frac{(3.07-2.15)^2}{(2.15-4.53)(4.53-3.07)} + \frac{(2.15-4.53)^2}{(4.53-3.07)(3.07-2.15)}$ is simplified to

Let: $x=4.53$, $y=3.07$ and $z=2.15$

The original expression becomes:

$\dfrac{(x-y)^{2}}{(y-z)(z-x)}+\dfrac{(y-z)^{2}}{(z-x)(x-y)}+\dfrac{(z-x)^{2}}{(x-y)(y-z)}$

$= \dfrac{(x-y)^{3}+ (y-z)^{3} +(z -x)^{3}}{(x-y)(y-z)(z-x)}$

Now, let $(x-y) = a , (y-z) = b$ and $(z-x) = c$

We observe, $(a + b + c) = x-y+y-z+z-x = 0$

Thus, $a^3 + b^3 + c^3 = 3abc$ (since $a+b+c = 0$)

Thus, the expression becomes:

$= \dfrac{a^3 + b^3 +c^3}{abc}= \dfrac{3abc}{abc} = 3$

Hence, the correct option is Option 4.