Can't solve the math related to power factor improvement

Image source - Fundamentals of Electric Circuit by Alexander & Sadiku, Practice problem 11.15.

I tried to solve the math in the following way.
$$$$S_{old}=140000\angle cos^{-1}(0.85) = 119000+j73749.576\\ S_{new}=140000\angle cos^{-1}(1.00) = 140000+j0.00 \;\;\;\;\;\;\;\;\;\; \\ So, Q_c = 73749.576 \\ And,\; C=\frac{Q_c}{2\pi f {V_{RMS}} ^2} = \frac{73749.576}{2\pi 60 (110) ^2} = 0.0161675\;F$$$$

Which is a wrong answer. Can anyone provide me the correct way?

• You were right, but I still find the text to be misleading: "Find the value [...] needed to correct a load of [...]" tells me that the load is 140 kVAr, and that needs to be corrected. Otherwise, I would have said "Find the value [...] that is used for a compensated power of [...]". It seems to convey a clearer message, but maybe it's just me (most possible). – a concerned citizen Jun 12 '20 at 22:06