$4096=2^{12}$ and multiplying a number in binary with $2^x$ means that we are shifting that number to the left by $x$ bits.
So, the number $4096 . 3$ will be $(11)_2$ shifted by $12$ bits to the left which will become $11000000000000$.
Same we can do with other numbers as well and will find that there is no overlapping $1's$. So total $1's = 1's$ in $3,15,5,3$.
So the answer is $10$.
Result =
$11000000000000$ $+$
$111100000000$ $+$
$1010000$ $+$
$11$
$=11111101010011$