Playing with Perfect Number!

Searching the first $10$ values for $k$ and $p$ using $Mathematica$

Select[Flatten@Table[2^k*3Prime@p,{k,10},{p,3,10}],1~DivisorSigma~#==3#&]

gave the correct answer $120$ , $672$

Since $n=2^k\cdot3p$ , the given equation gives $\sigma(2^k)\sigma(3)\sigma(p)=2^k\cdot3^2p$ $(2^{k+1}-1)\cdot4(p+1)=2^k\cdot9p$ $(2^{k+1}-1)(p+1)=2^{k-2}\cdot9p$ Because $\gcd(2^{k-2},2^{k+1}-1)=1$ so $2^{k-2}\mid p+1$ , in addition $9p$ is odd, so the number at both sides have precisely $2^{k-2}$ but not $2^{k-1}$ as its divisor.

It is sufficient to let $p+1=2^{k-2}q$ where $q$ is an odd number.

$(2^{k+1}-1)q=9(2^{k-2}q-1)$ $9=(2^{k-2}+1)q$

This equation sets limitation to the range of value of $q$ , reduce to only $\{1,3\}$ , $q\neq 9$ or $2^{k-2}=0$ ,

solving for $q=1,3$ respectively gives us $n=672,120$ , the answer is $672+120=\boxed{792}$ .

Also proving that $120$ and $672$ are the only two 3-perfect number of that form.