The three 2µf capacitors from the left are parallel. Therefore, effective capacitance in parallel = 6µf The 6µf, 2µf(opposite the 3µf), and the 3µf are in series. So effective/total capacitance in series = \(\frac{1}{C_T}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\) + \(\frac{1}{C_3}\) = \(\frac{1}{6}\) + \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{1 + 3 + 2}{6}\) = \(\frac{6}{6}\) C\(_T\) = 1µf