The two #2\ mu"F"# capacitors are in parallel, so they can be replaced by their equivalent capacitance of #4\ mu"F"#.
This leaves us with two parallel branches, each having two #4\ mu"F"# capacitors in series.
The capacitance of each branch is #(4\ mu"F"times 4\ mu"F")/(4\ mu"F"+4\ mu"F") = 2\ mu"F"#.
So, finally we have two #2\ mu"F"# capacitors in parallel, leading to an overall equivalent capacitance of #4\ mu"F"#
Note
For two capacitors #C_1# and #C_2# in series, the equivalent capacitance is #(C_1C_2)/(C_1+C_2)#, while the equivalent of parallel capacitors is #C_1+C_2#