Answer
${\bf 3}\;\rm \Omega$
Work Step by Step
We know that the force exerted by one wire on the other is given by
$$F=\dfrac{\mu_0 L_1I_1I_2}{2\pi d}$$
Assuming that the current in the left circuit is $I_1$ and the in the right circuit is $I_2$.
So we need to solve for $I_2$ to find $R$,
$$I_2=\dfrac{2\pi d F}{\mu_0 L_1I_1}\tag 1$$
And according to Ohm's law,
$\varepsilon_1=I_1R_1$, and $\varepsilon_2=I_2R$
So,
$I_1=\dfrac{\varepsilon_1}{R_1}$, and $I_2=\dfrac{\varepsilon_2}{R}$
Plug into (1),
$$\dfrac{\varepsilon_2}{R}=\dfrac{2\pi FdR_1}{\mu_0 L_1 \varepsilon_1} $$
Thus,
$$R=\dfrac{\mu_0 L_1 \varepsilon_1\varepsilon_2} {2\pi FdR_1}$$
Plug the known;
$$R=\dfrac{(4\pi\times 10^{-7})(0.1)(9) (9)}{2\pi (5\times 10^{-3})(5.4\times 10^{-5})(2)}$$
$$R=\color{red}{\bf 3}\;\rm \Omega$$
