Electric field in a certain region is given by → E = ( A x ^ 2 ^ i...

\overset{\to}{E} = \dfrac{A}{{x}^{2}}\overset{^}{i} + \dfrac{B}{{y}^{3}}\overset{^}{j}

Here, the units of $\dfrac{A}{{x}^{2}} and \dfrac{B}{{y}^{3}}$ is equal to unit of electric field.

The unit of electric field is $N{C}^{- 1}$.

So, the SI unit of $A$ is \left[\dfrac{A}{{x}^{2}}\right] = N{C}^{- 1} ⇒ \left[A\right] = N{m}^{2}{C}^{- 1}$</p><p>And the SI unit of$B$is \left[\dfrac{B}{{y}^{3}}\right] = N{C}^{- 1} ⇒ \left[B\right] = N{m}^{3}{C}^{- 1}$