The motion can be split into a component perpendicular to the wire ($v_y$, along $\hat{y}$) and a component parallel to it ($v_x$). Only the perpendicular component $v_y$ changes the flux through the loop and drives an EMF.

The magnetic field due to the long wire at distance $r$ is $B(r)=\frac{{\mu }_{0}I}{2\pi r}$. With the near side of the loop at distance $d$ and the far side at $d+a$ (where $a=4$ cm is the length perpendicular to the wire and $b=2$ cm is along the wire), the induced EMF as the loop moves with speed $v_y$ is

$\epsilon =(B(d)-B(d+a))b{v}_y=\frac{{\mu }_{0}Ib}{2\pi }\left(\frac{1}{d}-\frac{1}{d+a}\right)v_y$.

The induced current is $i=\epsilon /R$. Using ${\mu }_0=4\pi \times 10^{-7}$, $I=10$ A, $b=2\times 10^{-2}$ m, $d=4\times 10^{-2}$ m, $a=4\times 10^{-2}$ m, $R=0.1\Omega$, $i=10^{-5}$ A:

$10^{-5}=\frac{2\times 10^{-7}\cdot 10}{0.1}\left(\frac{1}{4}-\frac{1}{8}\right)\cdot 2v_y$,

which simplifies to $v_y=2$ m s${}^{-1}$.

Since the velocity direction is $\frac{\sqrt{3}}{2}\hat{x}+\frac{1}{2}\hat{y}$, we have $v_y=v/2$, hence $v=4$ m s${}^{-1}$.

$\therefore v=4{\text{m s}}^{-1}$.