Assigning E–Z configuration for alkenes with cyclic substituent groups
The compound is a tetra-substituted alkene with four different groups. Thus, it could be a (E)- or (Z)-ethene. To find the priority of the groups, you must follow CIP rules. The four groups are cyclobut-1-enyl, cyclobut-2-enyl, 2-methylcyclobutyl, and 1-methylcyclobutyl. Since it is an alkene (planner sp2–sp2 bond), we don't have to consider all four groups at once. What you have to do is consider two groups at once on each side of the double bond (say they are a and b):

On side a the first carbon of each cyclobutyl group (cyclobut-1-enyl and 1-methylcyclobutyl) has equal priority ({C,C,C} and {C,C,C}) as indicated in the image. However, the second carbon of cyclobut-1-enyl group has the priority {C,C,H}, while the second carbon of 1-methylcyclobutyl group has the lower priority {C,H,H} compared to that in cyclobut-1-enyl group. Therefore, between these two groups, cyclobut-1-enyl group has the higher priority.
On side b the first two carbons of each cyclobutyl group (cyclobut-2-enyl and 2-methylcyclobutyl) have equal priority: First carbons of both rings have {C,C,H} and the second carbons of both rings have {C,C,H} as well (see the image). Yet, the third carbon of cyclobut-2-enyl group has the priority {C,C,H} while the that of 2-methylcyclobutyl group has the lower priority {C,H,H} compared to cyclobut-1-enyl group. Therefore, between these two groups, the cyclobut-2-enyl group has a higher priority.
Now, you see higher priority groups of a and b sides are oriented on the same side of the double bond. Therefore, the compound is (Z)-isomer. The name is (Z)-1-(cyclobut-1-enyl)-2-(cyclobut-2-enyl)-1-(1-methylcyclobutyl)-2-(2-methylcyclobutyl)ethene.
Also, note that this molecule also has three chiral centers as indicated in the image. However, their stereochemistry is not given, so I neglect them.