Determine the equivalent capacitance for the group of capacitors in the drawing. Let all capacitors be the same where C = 14.0 µF.?

I redrew it for clarity

two on the right are in series, total is C/2

that is in parallel with the 3 in the middle, and they add

so total there is 3C + C/2 = (7/2)C

the final one on the left is in series, and we use the

C₁C₂/(C₁+C₂) formula, which gets us

C(7/2)C / (C+(7/2)C) = (7/2)C²/((9/2)C) = C(7/2)(2/9) = (7/9)C

or (7/9)14) = 10.9 µF

I'm labeling them a, b, c, d (across the middle of the picture) and e, f (across the bottom.

The combination of d and f is a series combination, whose equivalent capacitance is 7.0 uF. That is in parallel with a, b, and c; so the equivalent capacitance of a,b,c,d,f is 7 + 42 = 49.0 uF.

Finally, that combination is in series with e, so the equivalent capacitance for a,b,c,d,e,f is

1/(1/7 + 1/49) = 49/8 = 6.125 uF. Then use whatever round-off rule is customary in your class.