It only looks simple!

What a question!. I spent almost half an hour to solve this question and that also after a failed attempt.

Side View

First note that on left wire the force would be attractive and on the right wire the force will be repulsive. As we rotate the the loop through a small angle $\theta$ we will have a top view as shown below.

Top view

$Torque=\tau =\frac { l }{ 2 } ({ F }_{ 1 }sin{ \theta }_{ 1 }+{ F }_{ 2 }sin{ \theta }_{ 2 })\quad where\quad { \theta }_{ 1 }=\angle ABC\quad and\\ { \theta }_{ 2 }=\angle ADC.$

Also since $\theta$ is very small we have $AB\approx d$ and $AD\approx d+l$

First we will find ${ F }_{ 1 }$ and ${ F }_{ 2 }$

${ F }_{ 1 }={ B }_{ 1 }il=\frac { { \mu }_{ 0 }Iil }{ 2\pi d } ,{ F }_{ 2 }={ B }_{ 2 }il=\frac { { \mu }_{ 0 }Iil }{ 2\pi (d+l) }$

Now we will find ${ \theta }_{ 1 }$ and ${ \theta }_{ 2 }$

Applying sine rule in triangle ABC we have:

$\frac { AC }{ sin{ \theta }_{ 1 } } =\frac { AB }{ sin\theta } \quad \Rightarrow sin{ \theta }_{ 1 }=sin\theta (\frac { d+l/2 }{ d } ) \quad (i)$

Applying sine rule in triangle ADC we have :

$\frac { AC }{ sin{ \theta }_{ 2 } } =\frac { AD }{ sin(\pi -\theta ) } \quad \Rightarrow \frac { AC }{ sin{ \theta }_{ 2 } } =\frac { AD }{ sin\theta } \quad \Rightarrow sin{ \theta }_{ 2 }=sin\theta (\frac { d+l/2 }{ d+l } ) \quad (ii)$

Now putting the values of ${ F }_{ 1 },{ F }_{ 2 },sin{ \theta }_{ 1 },sin{ \theta }_{ 2 }$ we have :

$I\alpha =\tau =-(\frac { l }{ 2 } (\frac { { \mu }_{ 0 }Iil }{ 2\pi d } \frac { \theta (d+l/2) }{ d } +\frac { { \mu }_{ 0 }Iil }{ 2\pi (d+l) } \frac { \theta (d+l/2) }{ (d+l) } ))\quad (Since\quad sin\theta \approx \theta )$

One thing left to calculate is $I$

Using theorom of parellel axes we have :

${ I }_{ z }=4(\frac { m{ l }^{ 2 } }{ 4 } +\frac { m{ l }^{ 2 } }{ 12 } )=\frac { 4m{ l }^{ 2 } }{ 3 }$

Using therom of perpendicular axes we have:

${ I }_{ x }+{ I }_{ y }={ I }_{ z }$ and since ${ I }_{ x }={ I }_{ y }$ we have ${ I }_{ x }=\frac { { I }_{ z } }{ 2 } =\frac { 2m{ l }^{ 2 } }{ 3 }$

Putting this into our original equation we have :

$\alpha =-((\frac { { 3\mu }_{ 0 }Ii }{ 8\pi m } )(d+l/2)(\frac { 1 }{ { d }^{ 2 } } +\frac { 1 }{ { (d+l) }^{ 2 } } ))\theta$

$\Rightarrow \omega =\sqrt { (\frac { { 3\mu }_{ 0 }Ii }{ 8\pi m } )(d+l/2)(\frac { 1 }{ { d }^{ 2 } } +\frac { 1 }{ { (d+l) }^{ 2 } } ) }$ and $T=\frac { 2\pi }{ \omega }$

Put the values and we get $\boxed { T=45.88\quad sec }$

The torque along the Y axis is given by $U(o)Ii(a^2+b^2)2ab^2sin@/π((a^2-b^2)^2+4a^2b^2sin^2@)$ where $2b=L$ and a is the distance between the axes $(a=15cm)$ .For small displacements the $(sin@)^2$ term can be neglected and so the motion is almost Simple Harmonic.The moment of Inertia can be calculated easily and it is $I=2mL^2/3$ from parallel axis theorem.Solving $w=√3U(o)Iia/4mπ(4a^2+L^2)/(4a^2-L^2)$ and $T=2π/w$ = $45.885$ sec....Here $@$ represents small angular displacement.