Correct Answer - A
The electric potential `V (x,y,z) = 4x^(2)` volt
Now `vecE = - (hati (del V)/(delx) + hatj(delV)/(delx) + hatk(delV)/(delz))`
Now `(delV)/(delx) = 8x, (delV)/(dely) = 0` and `(delV)/(delz) = 0`
Hence `vecE = - 8x hati`, so at point `(1m, 0.2 m)`
`vecE = -8 hati` volt//metre or `8` along negative X-axis