Correct Answer - c
Magnetic field due to stright wire above O is zero
i.e `B_(1)=0(since,theta=0^(@))`
The magnetic field due to semi circular part
`B_(2)=(mu_(0)nI)/(2r)=(mu_(0)(1//2)I)/(2r)=(mu_(0)I)/(4r)`
The magnetic field due to lower straight portion
`B_(3)=(1)/(2)(mu_(0)I)/(2pir)(upward)`
Net magnetic field `B=B_(1)+B_(2)+B_(3)`
`=0+(mu_(0)I)/(4r)+(mu_(0)I)/(4pir)(upwards)`