Three current-carrying wires

We know that, ${ B }\quad =\quad \frac { { \mu }_{ 0 }I }{ 2\pi r }$ On the y-axis, ${ B }_{ M } = \frac {2 { \mu }_{ 0 }I }{ 2\pi r }$ and, ${ B }_{ N }= -\frac { { \mu }_{ 0 }I }{ 2\pi r }$ . (negative because it comes out of the plane and is opposite to ${B}_{M}$ )

The resultant will be ${ B }_{ M+N }\quad =\quad \frac { { \mu }_{ 0 }I }{ 2\pi r }$

The magnetic field produced by L has to counter this resultant. If $xI$ is the amount of current through L, then, ${ B }_{ L }\quad =\quad \frac { { \mu }_{ 0 }xI }{ 4\pi r }$ and hence, $\frac { { \mu }_{ 0 }xI }{ 4\pi r } = -\frac { { \mu }_{ 0 }I }{ 2\pi r }$ . Solving for x, we get $\boxed {x=-2}$ . Hence, current will be in -y direction and its intensity should be 2I