no... totally not. 1/2 F.e is not avg.. work...

what would be the meaning of the AVG..work done...

say.. you move a stone placed on road... to a distance... x

and the force you applied is... F which is constant...

so.. the work done by you is...W= F.x

avg... work would mean nothing...

BUT... if say the force is variable... (changing with time)

and again you move the stone to distance.. x..

then the work W = {integration of.. F.x }over the proper interval...

in this case.. if you wanted a crude answer.. the you could average out the force that was acting on the stone...

to... F/2 .. and you could have got a very.. ... sort of unreliable answer

and in your question.. you are given ... average... force... that was being acted upon the stone...