Answer
$$70.6 \ \mu F.$$
Work Step by Step
We are given:
$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$
We solve for the $C_2$ fraction:
$$\frac{1}{C_2}=\frac{1}{C}-\frac{1}{C_1}-\frac{1}{C_3}$$
Now, by substitution, we have
$$\frac{1}{C_2}=\frac{1}{25}-\frac{1}{75}-\frac{1}{80}=\frac{48}{1200}-\frac{16}{1200}-\frac{15}{1200}=\frac{17}{1200}$$
Hence, we get $$C_2=\frac{1200}{17}=70\frac{10}{17} \ \mu F\approx 70.6 \mu F.$$
