Let \(r_x\) and \(r_y\) be rates of working of Machine \(X\) and \(Y\) respectively

Machine \(X\) take 2 days more than Machine \(Y\),

\(t_x = t_y + 2 \Rightarrow t_y = t_x - 2\), where \(t_x\) and \(t_y\) are time taken by \(X\) and \(Y\) to produce \(w\) widgets alone respectively

Together, they produce \(\frac{5}{4}w\) widgets in \(3\) days,

\((r_x + r_y)3 = \frac{5}{4}w\)

\((\frac{w}{t_x} + \frac{w}{t_y})3 = \frac{5}{4}w\)

\((\frac{w}{t_x} + \frac{w}{t_x - 2})3 = \frac{5}{4}w\)

\((\frac{1}{t_x} + \frac{1}{t_x - 2})3 = \frac{5}{4}\)

\((\frac{t_x + t_x - 2}{t_x(t_x-2)})3 = \frac{5}{4}\)

\(\frac{6t_x - 6}{{t_x}^2 - 2t_x)} = \frac{5}{4}\)

\(24t_x - 24 = 5{t_x}^2 - 10t_x\)

\(5{t_x}^2 - 34t_x + 24 = 0\)

Using quadratic formula, \(x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}\)

\(t_x = \frac{-(-34) \pm\sqrt{(-34)^2 - 4*5*24}}{2*5} = \frac{34 \pm\sqrt{1156 - 480}}{10} = \frac{34 \pm26}{10} = \frac{34+26}{10} AND \frac{34-26}{10} = \textbf{6} AND \textbf{0.8}\)

0.8 days is not valid since \(X\) took 2 days more than \(Y\), so \(t_x > 2 days\)

Therefore, \(t_x = 6\) days

Time taken for Machine \(X\) to produce \(2w\) widgets alone = \(\frac{2w}{r_x} = \frac{2w}{\frac{w}{t_x}} = \frac{2w}{\frac{w}{6}} = \frac{12w}{w} = \textbf{12}\)

Hence, Answer is E