Step 1: Recall how nodal planes are assigned.
For any d-orbital written as $d_{ij}$, the four lobes sit in the $ij$ plane, and the two planes that do not contain both of those axes together are its nodal planes, where the electron density is exactly zero.
Step 2: Apply this to $3d_{xz}$.
Its lobes lie in the xz plane, so its nodal planes are the xy plane and the yz plane, meaning the electron density is zero in both of those planes.
Step 3: Screen the remaining statements quickly.
$3d_{xy}$ has nodal planes xz and yz, not the xy plane itself, so that option is wrong; $3d_{yz}$ has zero density in xy and xz, not in xz and xy together the way stated, so that option is wrong too; and 3d orbitals actually split under a magnetic field rather than staying degenerate, so that option fails as well.
Step 4: Conclude.
Only the statement about $3d_{xz}$ matches the nodal-plane rule.
Final answer: Option 2, the electron densities in the xy and yz planes are zero in the $3d_{xz}$ orbital.