Answer
(b), (d) and (e).
Work Step by Step
$E_{el}= k\frac{Q_{1}Q_{2}}{d}$
(a) When both charges are doubled,
$E_{el}=k\frac{2Q_{1}\times2Q_{2}}{d}=4\times k\frac{Q_{1}Q_{2}}{d}$
The magnitude of electrostatic attraction is 4 times greater.
(b) When we double one of the charges,
$E_{el}=k\frac{2Q_{1}Q_{2}}{d}=2\times k\frac{Q_{1}Q_{2}}{d}$
The magnitude of electrostatic attraction doubles.
(c) When distance between the charges are doubled,
$E_{el}=k\frac{Q_{1}Q_{2}}{2d}=\frac{1}{2}\times k\frac{Q_{1}Q_{2}}{d}$
(d) When distance is reduced to half,
$E_{el}=k\frac{Q_{1}Q_{2}}{\frac{1}{2}d}=2\times k\frac{Q_{1}Q_{2}}{d}$
The magnitude of electrostatic attraction doubles.
(e) When both charges are doubled and the distance between them are also doubled,
$E_{el}=k\frac{2Q_{1}\times2Q_{2}}{2d}=2\times k\frac{Q_{1}Q_{2}}{d}$
Again, the magnitude of electrostatic attraction doubles.
Thus the correct options are (b), (d) and (e).
