The current required to levitate the rod is given by

$I = dfrac{mg}{LB}$

If the length $L$ is doubled, the mass $m$ is also doubled but the ratio $dfrac{m}{L}$ remains same.

So the current needed to levitate the rod stays same.

The magnitude of the magnetic force on a wire of length $L$ carrying current $I$ lying perpendicular to the magnetic field $B$ is given by

$F = ILB$

Length of the wire is, $L = 2.15:m$.

Current on the wire is, $I = 0.695:A$.

Force acting on the wire is $F=0.25:N$.

So the magnetic field is given by

$B = dfrac{F}{IL} = dfrac{0.25}{0.695 times 2.15} = 0.167:T$

The magnitude of the magnetic force on a wire of length $L$ carrying current $I$ lying at an angle $theta$ with respect to a magnetic field $B$ is given by

$F= ILB sin theta$

Given data are

$I = 2.8:A$, $L = 2.25:m$. $B =0.88:T$ and $theta = 36^o$

Then

$F = 2.8 times 2.25 times 0.88 times sin 36^o = 3.26:N$