Answer
$2R$
Work Step by Step
The $R$-resistor is connected in parallel to the lower battery of $\varepsilon$, so its potential difference must be $\varepsilon$.
$$\Delta V_{R}=\varepsilon$$
The unknown resistor is connected in parallel to the two batteries. So it must have a potential difference of $2\varepsilon$.
$$\Delta V_{\rm unknown }=2\varepsilon$$
We know, according to Ohm's law, that
$$\Delta V =IR$$
Hence,
$$I=\dfrac{\Delta V }{R}$$
and since both resistors have the same current,
$$I=\dfrac{\Delta V_R }{R}=\dfrac{\Delta V_{\rm unknown } }{R_{\rm unknown }}$$
Hence,
$$R_{\rm unknown }=\dfrac{\Delta V_{\rm unknown } R}{\Delta V_R}$$
Plug the known;
$$R_{\rm unknown }=\dfrac{2\varepsilon R}{\varepsilon}$$
$$ \boxed{R_{\rm unknown }=2R}$$
