Correct Answer - C
The magnetic filed at the centre of the coil,
`B = (mu_(0))/(4pi) (2pi Ni)/(a)`
where `N` is number of turns, `i` current and `a`is radius of coil.
Given, `N = 25, D = 10 cm = 10xx10^(-2) km`
Radiius of the coil,
`a = (D)/(2) = (10xx10^(-2))/(2) m = 0.05m`
And current `i = 4A`
`:. B = 10^(-7) xx (2xx3.14xx25xx4)/(0.05)`
`B = 1.256xx10^(-3) T`