Answer
-0.28 V
Work Step by Step
$Sn^{2+}$ is reduced to $Sn$ and therefore the reaction is at the cathode.
$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=E^{\circ}_{Sn^{2+}/Sn}-E^{\circ}_{X^{2+}/X}$
$\implies E^{\circ}_{X^{2+}/X}=E^{\circ}_{Sn^{2+}/Sn}-E^{\circ}_{cell}=-0.14\,V-0.14\,V=-0.28\,V$
