Let x = rate at which water is flowing out from B

so Net rate =\(2 - x\)

SInce we know

Amount of work = Rate * time

or \(64 = (2-x) * 96\)

or \(64 = 192 - 96x\)

or\(x = \frac{128}{96} = \frac{4}{3}\)
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Rate of spout A = \(2\) g/hr
Time to fill \(64\) g drum by spout A = \(\frac{64}{2} = 32\) hr

Let time to fill the same drum by spout B = \(x\) hr

Since, they work together;
\(Time_{Total} = \frac{(T_A)(T_B)}{{T_A + T_B}}\)
\(96 = \frac{32x}{{32 + x}}\)
\((32)(3) + 3x = x\)
\(x = -48\)

Here, -ve sign means that the work done by spout B is opposite to that of spout A, which is absolutely correct.

Rate of spout B = \(\frac{Work}{Time_B} = \frac{64}{48} = \frac{4}{3}\) g/hr

Hence, option D
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