Suppose n=2^a3^b
2sigma(n) = (2^(a+1)-1)(3^(b+1)-1)
But sigma(n) = 3n, therefore =>
(2^(a+1)-1)(3^(b+1)-1) = 2^(a+1).3^(b+1)
By hcf arguments, therefore, only possible if:
2^(a+1) = 3^(b+1) - 1, and simultaneously,
3^(b+1) = 2^(a+1) - 1.
Which latter clearly not possible.